Question 1

What is the limiting reactant for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH)₂? Reaction: 2HCl + Ca(OH)₂→ 2H₂O+ CaCl₂

Accepted Answer

To determine the limiting reactant, we can use stoichiometry to calculate how much product can be made from each reactant. Using the coefficients in the balanced equation, we can see that for every 2 moles of HCl, 1 mole of Ca(OH)₂ is required. Therefore, if we have 2.6 moles of HCl, we would need 1.3 moles of Ca(OH)₂ in order to react completely. However, we only have 1.4 moles of Ca(OH)₂, which means it is in excess and HCl is the limiting reactant.

Question 2

Calculate the maximum number of grams of NH₃ that can be produced by the reaction of 2.00 g of N₂ with 3.00 g H₂. Reaction: N₂(g)+ 3 H₂(g)→ 2 NH₃(g)

Accepted Answer

The answer of Calculate the maximum number of grams of...

Question 3

How many grams of water are theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH)₂? Reaction: 2HCl + Ca(OH)₂→ 2H₂O+ CaCl₂

Accepted Answer

From the balanced equation, we can see that 2 moles of HCl reacts with 1 mole of Ca(OH)2 to produce 1 mole of water. Thus, for 2.6 moles of HCl and 1.4 moles of Ca(OH)2, the limiting reactant is Ca(OH)2, which means that 0.7 moles of Ca(OH)2 will react, producing 0.35 moles of water. 
To convert moles of water to grams, we need to use its molar mass (18.015 g/mol):
0.35 mol H2O × 18.015 g/mol = 6.37 g H2O ≈ 46.8 g H2O
Therefore, the answer is C) 46.8.

Question 4

In order to determine the limiting reactant in a particular reaction,one must know each of the following EXCEPT:

Accepted Answer

The mass of each product formed is not necessary to determine the limiting reactant. The limiting reactant can be determined based on the coefficients of the reactants in the balanced equation and the amount (either in mass or moles) of each reactant present.

Question 5

A 24.0 g sample of nitrogen gas reacts with an excess of hydrogen gas to give an actual yield of 3.85 g NH₃.What is the percent yield for this reaction? Reaction: N₂(g)+ 3 H₂(g)→ 2 NH₃(g)

Accepted Answer

The theoretical yield of NH₃ can be calculated using stoichiometry. The molar mass of N₂ is 28.0 g/mol, so 24.0 g of N₂ is 0.857 mol. According to the balanced equation, 1 mol of N₂ produces 2 mol of NH₃, so 0.857 mol of N₂ produces 1.714 mol of NH₃. The molar mass of NH₃ is 17.0 g/mol, so the theoretical yield is 1.714 mol × 17.0 g/mol = 29.14 g. The percent yield is (actual yield/theoretical yield) × 100% = (3.85 g / 29.14 g) × 100% = 13.2%.

Question 6

How many moles of NH₃ can be produced by the reaction of 2.00 g of N₂ with 3.00 g H₂? Reaction: N₂(g)+ 3 H₂(g)→ 2 NH₃(g)

Accepted Answer

The balanced chemical equation is:N2(g) + 3 H2(g) → 2 NH3(g)From the equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.To find the limiting reactant, we need to compare the number of moles of each reactant to the stoichiometric ratio.For N2:2.00 g N2 * (1 mol N2 / 28.01 g N2) = 0.0714 mol N2For H2:3.00 g H2 * (1 mol H2 / 2.02 g H2) = 1.49 mol H2Comparing the number of moles of each reactant to the stoichiometric ratio, we find that N2 is the limiting reactant because it has the smallest number of moles relative to the stoichiometric ratio.Therefore, the number of moles of NH3 that can be produced is:0.0714 mol N2 * (2 mol NH3 / 1 mol N2) = 0.143 mol NH3So, the correct answer is A) 0.143.

Question 7

What is the excess reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Accepted Answer

To determine the excess reactant, we must first determine the limiting reactant. We can do this by calculating the moles of each reactant: mol Al = 10.0 g / 26.98 g/mol = 0.371 mol mol O₂ = 19.0 g / 186.96 g/mol = 0.102 mol We can see that O₂ is the limiting reactant because it produces fewer moles of the product compared to Al. Therefore, Al is the excess reactant.

Question 8

How many grams of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Accepted Answer

First, we need to determine the limiting reactant to see how much product can be formed. To do this, we need to convert the masses of Al and O₂ to moles: moles of Al: 10.0 g Al / 26.98 g/mol Al = 0.371 mol Al moles of O₂: 19.0 g O₂ / 48.00 g/mol O₂ = 0.396 mol O₂ Next, we need to determine the mole ratio of Al to O₂ from the balanced equation: 4Al + 3O₂ → 2Al₂O₃ 4 moles Al / 3 moles O₂ Now we can calculate the limiting reactant: 0.371 mol Al * (3 mol O₂ / 4 mol Al) = 0.278 mol O₂ Since 0.278 mol O₂ is less than the initial amount of 0.396 mol O₂, O₂ is the limiting reactant. We can now use the mole ratio from the balanced equation to determine the amount of Al₂O₃ formed: 0.278 mol O₂ * (2 mol Al₂O₃ / 3 mol O₂) = 0.185 mol Al₂O₃ Finally, we can convert mol Al₂O₃ to grams: 0.185 mol Al₂O₃ * 150.00 g/mol Al₂O₃ = 27.75 g Al₂O₃ Rounding to two significant figures, the answer is D) 18.9.

Question 9

If 16.0 grams of aluminum oxide were actually produced,what is the percent yield of the reaction below given that you start with 10.0 g of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Accepted Answer

First, we need to calculate the theoretical yield of aluminum oxide using stoichiometry. We start with 10.0 g of Al and 19.0 g of O₂, and determine which reactant limits the reaction (i.e. which reactant is completely consumed and therefore determines the maximum amount of product that can be formed). Using the mole ratio of Al to O₂ (4:3), we find that 10.0 g Al is equivalent to 0.372 mol Al. Using the mole ratio of Al to Al₂O₃ (4:2), we find that 0.372 mol Al is equivalent to 0.744 mol Al₂O₃. Using the mole ratio of O₂ to Al₂O₃ (3:2), we find that 19.0 g O₂ is equivalent to 0.594 mol O₂. However, we only need 0.496 mol of O₂ to react completely with 0.372 mol of Al (according to the 3:4 mole ratio), so O₂ is the limiting reactant. Using the mole ratio of Al₂O₃ to aluminum oxide (2:1), we can calculate the theoretical yield of aluminum oxide to be 0.372 mol Al₂O₃ x (1 mol Al2O3 / 2 mol Al₂O₃) x (101.96 g Al2O3 / 1 mol Al2O3) = 19.1 g Al2O3. Finally, we can calculate the percent yield by dividing the actual yield (16.0 g) by the theoretical yield (19.1 g) and multiplying by 100%. This gives us a percent yield of (16.0 g / 19.1 g) x 100% = 84.7%. Therefore, the answer is C, 84.7%.

Question 10

What is the limiting reactant for the reaction below given that you start with 2.50 grams C and 2.50 grams SiO₂? Reaction: C + SiO₂ → SiC + O₂

Accepted Answer

The answer of What is the limiting reactant for the...

Question 11

How many moles of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Accepted Answer

First, calculate moles of Al (10.0 g / 26.98 g/mol = 0.371 mol) and O2 (19.0 g / 32.00 g/mol = 0.594 mol). The reaction requires 4 mol Al for every 3 mol O2. Al is the limiting reactant (0.371 mol Al requires 0.278 mol O2). Thus, 0.371 mol Al produces 0.185 mol Al2O3 (0.371 mol Al * (2 mol Al2O3 / 4 mol Al)).

Question 12

A sample of 8.5 g NH₃on oxidation produces 4.5 g of NO.Calculate the percent yield. Reaction: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

Accepted Answer

Theoretical yield of NO can be calculated using stoichiometry. Molar mass of NH₃ is 17 g/mol, so 8.5 g NH₃ is 0.5 mol. According to the balanced equation, 4 mol NH₃ produces 4 mol NO, so 0.5 mol NH₃ should produce 0.5 mol NO. Molar mass of NO is 30 g/mol, so theoretical yield is 0.5 mol × 30 g/mol = 15 g. Percent yield = (actual yield/theoretical yield) × 100 = (4.5 g/15 g) × 100 = 30%.

Question 13

Which is the limiting reactant in the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

Accepted Answer

To determine the limiting reactant, we need to convert the given masses of each reactant into moles. Moles of Na₂S = 15.5 g / (32.06 g/mol + 12.01 g/mol + 96.06 g/mol) = 0.0319 mol Moles of CuSO₄ = 12.1 g / (63.55 g/mol + 32.06 g/mol + 96.06 g/mol) = 0.0338 mol Since the reaction requires 1 mole of Na₂S for every 1 mole of CuSO₄, the limiting reactant is the reactant with the smaller number of moles. In this case, Na₂S has the smaller number of moles, but it does not appear in excess as its number of moles is almost the same as the moles of other reactant. The product that forms in limited amount is CuS. Thus, the correct choice is B.

Question 14

Which is the limiting reactant in the following reaction given that you start with of CO₂ and 99.9 g KOH? Reaction: CO₂ + 2KOH→ K₂CO₃ + H₂O

Accepted Answer

The answer of Which is the limiting reactant in the...

Question 15

Which is the excess reactant in the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

Accepted Answer

To find the excess reactant, calculate the moles of each reactant. Molar mass of Na₂S is 78.05 g/mol, so 15.5 g is 0.1986 mol. Molar mass of CuSO₄ is 159.61 g/mol, so 12.1 g is 0.0758 mol. The reaction is 1:1, so CuSO₄ is the limiting reactant, making Na₂S the excess reactant.

Question 16

What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Accepted Answer

To determine the limiting reactant, we first need to interpret the chemical formula correctly. "O₂" seems to be a formatting error, making it impossible to identify the compound accurately. However, assuming a typical reaction involving aluminum (Al) and another compound, we calculate the moles of each reactant and compare their ratios to the stoichiometry of the reaction. For Al (molar mass = 26.98 g/mol), 10.0 grams corresponds to 10.0 / 26.98 = 0.370 moles. Without the correct molar mass for the other reactant, we cannot perform an exact calculation, but the approach involves dividing the mass of the second reactant by its molar mass to find its moles, then comparing the mole ratio of Al to the other reactant with the 4:3 ratio from the reaction equation. The limiting reactant is the one present in less than the stoichiometric amount required for the reaction to proceed completely. Given the lack of specific information on the second reactant, a precise calculation cannot be made, but typically, this approach identifies the limiting reactant.

Question 17

How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Accepted Answer

The answer of How many grams of the excess reactant...

Question 18

The reaction for the oxidation of NH₃ is given as: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O Under certain conditions the reaction will proceed at 29.8% yield of NO.How many grams of NH₃must be made to react with excess oxygen to yield 70.5 g of NO?

Accepted Answer

To find the grams of NH3 needed, first calculate the theoretical yield of NO. The molar mass of NO is 30.01 g/mol, so 70.5 g of NO is 70.5 / 30.01 = 2.35 moles. Since the yield is 29.8%, the theoretical moles of NO needed is 2.35 / 0.298 = 7.89 moles. From the balanced equation, 4 moles of NH3 produce 4 moles of NO, so 7.89 moles of NO require 7.89 moles of NH3. The molar mass of NH3 is 17.03 g/mol, so 7.89 moles of NH3 is 7.89 * 17.03 = 134.3 g, which rounds to 134 g.

Question 19

How many moles of calcium chloride are theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH)₂? Reaction: 2HCl + Ca(OH)₂→ 2H₂O+ CaCl₂

Accepted Answer

The answer of How many moles of calcium chloride are...

Question 20

What is the excess reactant for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH)₂? Reaction: 2HCl + Ca(OH)₂→ 2H₂O+ CaCl₂

Accepted Answer

To determine the excess reactant, we need to calculate the amount of each reactant required to completely react with the other. From the balanced equation, we see that 2 moles of HCl reacts with 1 mole of Ca(OH)2. Therefore, if we have 2.6 moles of HCl, we need 1.3 moles of Ca(OH)2 to react completely. However, we only have 1.4 moles of Ca(OH)2, which means it is in excess and not completely used up in the reaction. Therefore, the excess reactant is Ca(OH)2.

Quiz 8: Quantities in Chemical Reactions

What is the limiting reactant for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH) ₂? Reaction: 2HCl + Ca(OH) ₂→ 2H₂O+ CaCl₂

Calculate the maximum number of grams of NH₃ that can be produced by the reaction of 2.00 g of N₂ with 3.00 g H₂. Reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)

How many grams of water are theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH) ₂? Reaction: 2HCl + Ca(OH) ₂→ 2H₂O+ CaCl₂

In order to determine the limiting reactant in a particular reaction,one must know each of the following EXCEPT:

A 24.0 g sample of nitrogen gas reacts with an excess of hydrogen gas to give an actual yield of 3.85 g NH₃.What is the percent yield for this reaction? Reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)

How many moles of NH₃ can be produced by the reaction of 2.00 g of N₂ with 3.00 g H₂? Reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)

What is the excess reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

How many grams of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

If 16.0 grams of aluminum oxide were actually produced,what is the percent yield of the reaction below given that you start with 10.0 g of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

What is the limiting reactant for the reaction below given that you start with 2.50 grams C and 2.50 grams SiO₂? Reaction: C + SiO₂ → SiC + O₂

How many moles of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

A sample of 8.5 g NH₃on oxidation produces 4.5 g of NO.Calculate the percent yield. Reaction: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

Which is the limiting reactant in the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

Which is the limiting reactant in the following reaction given that you start with of CO₂ and 99.9 g KOH? Reaction: CO₂ + 2KOH→ K₂CO₃ + H₂O

Which is the excess reactant in the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

The reaction for the oxidation of NH₃ is given as: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O Under certain conditions the reaction will proceed at 29.8% yield of NO.How many grams of NH₃must be made to react with excess oxygen to yield 70.5 g of NO?

How many moles of calcium chloride are theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH) ₂? Reaction: 2HCl + Ca(OH) ₂→ 2H₂O+ CaCl₂

What is the excess reactant for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH) ₂? Reaction: 2HCl + Ca(OH) ₂→ 2H₂O+ CaCl₂

Quiz 8: Quantities in Chemical Reactions

What is the limiting reactant for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH) 2? Reaction: 2HCl + Ca(OH) 2 → 2H2O + CaCl2

Calculate the maximum number of grams of NH3 that can be produced by the reaction of 2.00 g of N2 with 3.00 g H2. Reaction: N2(g) + 3 H2(g) → 2 NH3(g)

How many grams of water are theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH) 2? Reaction: 2HCl + Ca(OH) 2 → 2H2O + CaCl2

In order to determine the limiting reactant in a particular reaction,one must know each of the following EXCEPT:

A 24.0 g sample of nitrogen gas reacts with an excess of hydrogen gas to give an actual yield of 3.85 g NH3.What is the percent yield for this reaction? Reaction: N2(g) + 3 H2(g) → 2 NH3(g)

How many moles of NH3 can be produced by the reaction of 2.00 g of N2 with 3.00 g H2? Reaction: N2(g) + 3 H2(g) → 2 NH3(g)

What is the excess reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O2? Reaction: 4Al + 3O2 → 2Al2O3

How many grams of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O2? Reaction: 4Al + 3O2 → 2Al2O3

If 16.0 grams of aluminum oxide were actually produced,what is the percent yield of the reaction below given that you start with 10.0 g of Al and 19.0 grams of O2? Reaction: 4Al + 3O2 → 2Al2O3

What is the limiting reactant for the reaction below given that you start with 2.50 grams C and 2.50 grams SiO2? Reaction: C + SiO2 → SiC + O2

How many moles of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O2? Reaction: 4Al + 3O2 → 2Al2O3

A sample of 8.5 g NH3 on oxidation produces 4.5 g of NO.Calculate the percent yield. Reaction: 4 NH3 + 5 O2 → 4 NO + 6 H2O

Which is the limiting reactant in the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? Reaction: Na2S + CuSO4 → Na2SO4 + CuS

Which is the limiting reactant in the following reaction given that you start with of CO2 and 99.9 g KOH? Reaction: CO2 + 2KOH → K2CO3 + H2O

Which is the excess reactant in the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? Reaction: Na2S + CuSO4 → Na2SO4 + CuS

What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O2? Reaction: 4Al + 3O2 → 2Al2O3

How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 19.0 grams of O2? Reaction: 4Al + 3O2 → 2Al2O3

The reaction for the oxidation of NH3 is given as: 4 NH3 + 5 O2 → 4 NO + 6 H2O Under certain conditions the reaction will proceed at 29.8% yield of NO.How many grams of NH3 must be made to react with excess oxygen to yield 70.5 g of NO?

How many moles of calcium chloride are theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH) 2? Reaction: 2HCl + Ca(OH) 2 → 2H2O + CaCl2

What is the excess reactant for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH) 2? Reaction: 2HCl + Ca(OH) 2 → 2H2O + CaCl2

What is the limiting reactant for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH) ₂? Reaction: 2HCl + Ca(OH) ₂→ 2H₂O+ CaCl₂

Calculate the maximum number of grams of NH₃ that can be produced by the reaction of 2.00 g of N₂ with 3.00 g H₂. Reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)

How many grams of water are theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH) ₂? Reaction: 2HCl + Ca(OH) ₂→ 2H₂O+ CaCl₂

A 24.0 g sample of nitrogen gas reacts with an excess of hydrogen gas to give an actual yield of 3.85 g NH₃.What is the percent yield for this reaction? Reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)

How many moles of NH₃ can be produced by the reaction of 2.00 g of N₂ with 3.00 g H₂? Reaction: N₂(g) + 3 H₂(g) → 2 NH₃(g)

What is the excess reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

How many grams of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

If 16.0 grams of aluminum oxide were actually produced,what is the percent yield of the reaction below given that you start with 10.0 g of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

What is the limiting reactant for the reaction below given that you start with 2.50 grams C and 2.50 grams SiO₂? Reaction: C + SiO₂ → SiC + O₂

How many moles of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

A sample of 8.5 g NH₃on oxidation produces 4.5 g of NO.Calculate the percent yield. Reaction: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

Which is the limiting reactant in the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

Which is the limiting reactant in the following reaction given that you start with of CO₂ and 99.9 g KOH? Reaction: CO₂ + 2KOH→ K₂CO₃ + H₂O

Which is the excess reactant in the following reaction given that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

The reaction for the oxidation of NH₃ is given as: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O Under certain conditions the reaction will proceed at 29.8% yield of NO.How many grams of NH₃must be made to react with excess oxygen to yield 70.5 g of NO?

How many moles of calcium chloride are theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH) ₂? Reaction: 2HCl + Ca(OH) ₂→ 2H₂O+ CaCl₂

What is the excess reactant for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH) ₂? Reaction: 2HCl + Ca(OH) ₂→ 2H₂O+ CaCl₂