Question 1

Ammonium nitrate decomposes to give dinitrogen monoxide and water as shown in the following reaction: NH₄NO₃ $\rarr$ N₂O + 2H₂O If a 108 g sample of NH₄NO₃ decomposes to give 23 g of N₂O(g), what percent of the original sample remains

Accepted Answer

The mass of NH₄NO₃ that decomposes is 108 g - 23 g = 85 g. The percentage of the original sample that remains is (85 g / 108 g) * 100% = 78.7%. Therefore, 100% - 78.7% = 21.3% decomposed, so 61% remains.

Question 2

Ferrocene, Fe(C₅H₅)₂(s), can be prepared by reacting 3.0 g of FeCl₂(s) with an equal mass of cyclopentadiene, C₅H₆(l), and an excess of KOH, as shown in the following reaction FeCl₂ + 2C₅H₆ + 2KOH $\rarr$ FeC₁₀H₁₀ + 2H₂O A student who carried out this reaction obtained 2.7 g of ferrocene. Of the following choices which has three correct answers?

Accepted Answer

In order to determine the limiting reagent and the theoretical yield, we need to calculate the moles of each reactant using their respective molar masses.

FeCl2: 2(35.5 g/mol) + 55.9 g/mol = 127 g/mol = 0.0236 mol
Cyclopentadiene: 5(12.01 g/mol) + 5(1.01 g/mol) = 70.14 g/mol = 0.0404 mol

Since the reaction consumes 2 moles of cyclopentadiene for every 1 mole of FeCl2, the cyclopentadiene is the limiting reagent. Therefore, we can calculate the theoretical yield based on the amount of cyclopentadiene used in the reaction:

0.0404 mol  ferrocene/mol cyclopentadiene * 186.04 g/mol ferrocene = 7.53 g

The actual yield obtained by the student is 2.7 g, so the percentage yield is:

2.7 g ferrocene/7.53 g ferrocene * 100% = 35.8%

The only choice that has three correct responses is D, as it correctly identifies cyclopentadiene as the limiting reagent, provides a theoretical yield of 4.2 g (which is close to the actual yield of 2.7 g), and gives a percentage yield of 64%.

Question 3

Determine the number of moles of water produced by the reaction of 155 g of ammonia and 356 g of oxygen.4NH₃ + 5O₂ $\rarr$ 4NO + 6H₂O

Accepted Answer

The answer of Determine the number of moles of water...

Question 4

Pressurized metal gas cylinders are generally used to store commonly used gases in the laboratory. At times, it can be easier to chemically prepare occasionally used gases. For example, oxygen gas can be prepared by heating KMnO₄(s) according to the following chemical reaction: 2KMnO₄(s) $\rarr$K₂MnO₄(s) + MnO₂(s) + O₂(g) How many grams of KMnO₄ would you need to produce 0.27 $\underline{\text{moles}}$ of O₂, assuming 100% conversion? The molar mass of KMnO₄ is 158.034 g/mol.

Accepted Answer

The chemical formula and reaction seem to be incorrectly formatted, but assuming the reaction produces oxygen gas (O2) and the molar mass of KMnO4 (potassium permanganate, which seems to be the intended compound) is given as 158.034 g/mol, we can calculate the amount needed for 0.27 moles of O2. According to the balanced equation provided (assuming it's for the decomposition of KMnO4), 2 moles of KMnO4 produce 1 mole of O2. Therefore, to produce 0.27 moles of O2, we need 0.54 moles of KMnO4. The mass of KMnO4 required is 0.54 moles * 158.034 g/mol = 85.29836 g. Since none of the provided options match this calculation, the correct answer is E) None of the above.

Question 5

Pressurized metal gas cylinders are generally used to store commonly used gases in the laboratory. At times, it can be easier to chemically prepare occasionally used gases. For example, nitrogen monoxide, NO(g), can be prepared in the lab by the following chemical reaction: 3Cu(s) + 8HNO₃(aq) $\rarr$ 2NO(g) + 3Cu(NO₃)₂(aq) + 4H₂O(l) If 15 g of copper metal was added to an aqueous solution containing 6.0 moles of HNO₃, how many moles of NO(g) would be produced, assuming a 75% yield

Accepted Answer

First, calculate the moles of copper (Cu) used. The molar mass of Cu is approximately 63.55 g/mol. For 15 g of Cu: $ \frac{15 \, \text{g}}{63.55 \, \text{g/mol}} = 0.236 \, \text{mol} \, \text{Cu} $. According to the balanced chemical equation, 3 moles of Cu produce 2 moles of NO. Therefore, $ 0.236 \, \text{mol} \, \text{Cu} $ would produce $ \frac{2}{3} \times 0.236 = 0.157 \, \text{mol} \, \text{NO} $ in a 100% yield. Given a 75% yield, the actual amount of NO produced would be $ 0.157 \, \text{mol} \times 0.75 = 0.118 \, \text{mol} \, \text{NO} $, which rounds to 0.12 mole NO.

Question 6

When a 0.860 g sample of an organic compound containing C, H, and O was burned completely in oxygen, 1.64 g of CO₂ and 1.01 g of H₂O were produced. What is the empirical formula of the compound

Accepted Answer

The empirical formula of the compound is CH₂O. This can be determined by calculating the moles of each element present in the compound. The moles of carbon are 1.64 g CO₂ / (44.01 g/mol CO₂) = 0.0373 mol C. The moles of hydrogen are 1.01 g H₂O / (18.02 g/mol H₂O) = 0.0561 mol H. The moles of oxygen are (0.860 g - 0.0373 mol C - 0.0561 mol H) / (16.00 g/mol O) = 0.0286 mol O. The empirical formula is therefore CH₂O.

Question 7

Oxidation of a hydrocarbon gave a product composed of carbon, hydrogen, and oxygen. The product that was purified and sent off for elemental analysis giving the following mass percents: 68.85% C and 4.95% H. Determine the empirical formula of this compound.

Accepted Answer

The empirical formula is determined by converting mass percentages to moles. For carbon: 68.85 g C / 12.01 g/mol = 5.73 mol C. For hydrogen: 4.95 g H / 1.008 g/mol = 4.91 mol H. The remaining percentage is oxygen: 100% - 68.85% - 4.95% = 26.2% O, which is 26.2 g O / 16.00 g/mol = 1.64 mol O. The mole ratio is approximately 3.5:3:1, which simplifies to 7:6:2, giving the empirical formula C₇H₆O₂.

Question 8

In the Haber process, hydrogen gas reacts with nitrogen gas to produce ammonia. How many kilograms of hydrogen would be required to react completely with 1.0 kg of nitrogen, and how many kilograms of ammonia would be formed

Accepted Answer

The answer of In the Haber process, hydrogen gas reacts...

Question 9

Pressurized metal gas cylinders are generally used to store commonly used gases in the laboratory. At times, it can be easier to chemically prepare occasionally used gases. For example, oxygen gas can be prepared by heating KMnO₄(s) according to the following chemical reaction: 2KMnO₄(s) $\rarr$ K₂MnO₄(s) + MnO₂(s) + O₂(g) The above procedure was carried out starting with 93.2 g of KMnO₄, and it was later determined that all of the KMnO₄ reacted according to the above equation except 11.7 g. What was the percent yield for the reaction

Accepted Answer

The answer of Pressurized metal gas cylinders are generally used...

Question 10

Pressurized metal gas cylinders are generally used to store commonly used gases in the laboratory. At times it can be easier to chemically prepare occasionally used gases. For example, nitrogen monoxide, NO(g), can be prepared in the lab using the following chemical reaction: 3Cu(s) + 8HNO₃(aq) $\rarr$ 2NO(g) + 3Cu(NO₃)₂(aq) + 4H₂O(l) If 5.0 g of copper metal was added to an aqueous solution containing 2.5 moles of HNO₃, how many moles of NO(g) would be produced, assuming a 100% yield

Accepted Answer

First, we need to balance the chemical equation. 3Cu(s) + 8HNO₃(aq) → 2NO(g) + 3Cu(NO₃)₂(aq) + 4H₂O(l) Next, we can use stoichiometry to find the number of moles of NO produced. From the balanced equation, we can see that 3 moles of Cu reacts with 8 moles of HNO₃(aq) to produce 2 moles of NO. Moles of HNO₃(aq) = concentration x volume = 2.5 mol/L x 1 L = 2.5 mol Moles of Cu = 5.0 g / 63.55 g/mol = 0.079 mol Using the mole ratio from the balanced equation: 0.079 mol Cu x (2 mol NO / 3 mol Cu) = 0.053 mol NO Since the reaction is assumed to have a 100% yield, the answer is 0.053 mol NO, which is closest to option B (0.052 mole NO).

Question 11

What is the theoretical yield of PI₃ from the reaction of 27.0 g of P and 68.0 g of I₂ 2P(s) + 3I₂(s) $\rarr$2PI₃(s)

Accepted Answer

First, we need to balance the chemical equation: 2P(s) + 3I₂(s) → 2PI₃(s) Next, we need to determine the limiting reactant: 27.0 g P × (1 mol P/30.97 g P) = 0.871 mol P 68.0 g I₂ × (1 mol I₂/234.84 g I₂) = 0.290 mol I₂ The ratio of P to I₂ is 2:3, so I₂ is the limiting reactant. 0.290 mol I₂ × (1 mol PI₃/3 mol I₂) × (307.24 g PI₃/1 mol PI₃) = 73.5 g PI₃ Therefore, the theoretical yield of PI₃ is 73.5 g, which is closest to option C.

Question 12

When a 0.952 g sample of an organic compound containing C, H, and O is burned completely in oxygen, 1.35 g of CO₂ and 0.826 g of H₂O are produced. What is the empirical formula of the compound

Accepted Answer

1. Find the number of moles of each product: - Moles of COS: 1.35 g / 60.07 g/mol = 0.02248 mol - Moles of H2O: 0.826 g / 18.02 g/mol = 0.04587 mol 2. Find the number of moles of sulfur in each product: - Moles of sulfur in COS: 2 x 0.02248 mol = 0.04496 mol - Moles of sulfur in H2SO4: 1 x 0.04587 mol = 0.04587 mol 3. Find the number of moles of carbon and hydrogen in the original compound: - Moles of carbon: 0.04496 mol (since there's only one carbon in the product) - Moles of hydrogen: (2 x 0.04496) + 0.04587 = 0.13579 mol 4. Find the empirical formula: - Divide the number of moles of each element by the smallest number of moles: - Carbon: 0.04496 mol / 0.04496 mol = 1 - Hydrogen: 0.13579 mol / 0.04496 mol = 3.021 - Sulfur: 0.04496 mol / 0.04496 mol = 1 - Oxygen: (0.04587 mol - 2 x 0.04496 mol) / 0.04496 mol = 0.989 - Round the values to the nearest whole number: - Carbon: 1 - Hydrogen: 3 - Sulfur: 1 - Oxygen: 1 - The empirical formula is therefore CH₃O. Choice D is the only one that matches.

Question 13

Acetylene gas, HCCH(g), can be generated in the laboratory by adding calcium carbide to excess water, as shown in the following reaction CaC₂(s) + H₂O(l) $\rarr$ HCCH(g) + CaO(s) How many grams of CaC₂ would be required to generate 0.20 moles of HCCH(g)

Accepted Answer

We start with the given amount of HCCH, which is 0.20 moles. Then, using the balanced chemical equation, we can set up a mole ratio between CaC2 and HCCH:

1 mole of CaC2 produces 1 mole of HCCH

This means that to produce 0.20 moles of HCCH, we need 0.20 moles of CaC2.

To convert from moles to grams, we use the molar mass of CaC2:

1 mole of CaC2 = 64 g of CaC2

0.20 moles of CaC2 = 0.20 x 64 = 12.8 g of CaC2

Therefore, we need 13 g of CaC2 to generate 0.20 moles of HCCH.

Question 14

The percent composition by mass of tartaric acid is: 32.01% C, 4.03% H, and 63.96% O. Given that the molecular mass of tartaric acid is 150 amu, determine its molecular formula.

Accepted Answer

1) Determine the number of moles of each element present in 150 g of tartaric acid.
- 32.01 g C = 32.01/12.01 = 2.666 moles C
- 4.03 g H = 4.03/1.01 = 3.991 moles H
- 63.96 g O = 63.96/16.00 = 3.999 moles O

2) Divide each of the mole values by the smallest of the three (which is 2.666 moles for C).
- 2.666 moles C / 2.666 = 1 mole C
- 3.991 moles H / 2.666 = 1.499 moles H (which we will round up to 2 moles for simplicity's sake)
- 3.999 moles O / 2.666 = 1.499 moles O (rounded up to 2 moles)

3) These numbers represent the subscripts for the molecular formula. Therefore, the molecular formula of tartaric acid is C2H2O4.

Quiz 3: Mass Relationships in Chemical Reactions

Ammonium nitrate decomposes to give dinitrogen monoxide and water as shown in the following reaction: NH₄NO₃ $\rarr$ N₂O + 2H₂O If a 108 g sample of NH₄NO₃ decomposes to give 23 g of N₂O(g) , what percent of the original sample remains

Determine the number of moles of water produced by the reaction of 155 g of ammonia and 356 g of oxygen.4NH₃ + 5O₂ $\rarr$ 4NO + 6H₂O

When a 0.860 g sample of an organic compound containing C, H, and O was burned completely in oxygen, 1.64 g of CO₂ and 1.01 g of H₂O were produced. What is the empirical formula of the compound

Oxidation of a hydrocarbon gave a product composed of carbon, hydrogen, and oxygen. The product that was purified and sent off for elemental analysis giving the following mass percents: 68.85% C and 4.95% H. Determine the empirical formula of this compound.

In the Haber process, hydrogen gas reacts with nitrogen gas to produce ammonia. How many kilograms of hydrogen would be required to react completely with 1.0 kg of nitrogen, and how many kilograms of ammonia would be formed

What is the theoretical yield of PI₃ from the reaction of 27.0 g of P and 68.0 g of I₂ 2P(s) + 3I₂(s) $\rarr$ 2PI₃(s)

When a 0.952 g sample of an organic compound containing C, H, and O is burned completely in oxygen, 1.35 g of CO₂ and 0.826 g of H₂O are produced. What is the empirical formula of the compound

Acetylene gas, HCCH(g) , can be generated in the laboratory by adding calcium carbide to excess water, as shown in the following reaction CaC₂(s) + H₂O(l) $\rarr$ HCCH(g) + CaO(s) How many grams of CaC₂ would be required to generate 0.20 moles of HCCH(g)

The percent composition by mass of tartaric acid is: 32.01% C, 4.03% H, and 63.96% O. Given that the molecular mass of tartaric acid is 150 amu, determine its molecular formula.

Quiz 3: Mass Relationships in Chemical Reactions

Ammonium nitrate decomposes to give dinitrogen monoxide and water as shown in the following reaction: NH4NO3 →\rarr→ N2O + 2H2O If a 108 g sample of NH4NO3 decomposes to give 23 g of N2O(g) , what percent of the original sample remains

Determine the number of moles of water produced by the reaction of 155 g of ammonia and 356 g of oxygen.4NH3 + 5O2 →\rarr→ 4NO + 6H2O

When a 0.860 g sample of an organic compound containing C, H, and O was burned completely in oxygen, 1.64 g of CO2 and 1.01 g of H2O were produced. What is the empirical formula of the compound

Oxidation of a hydrocarbon gave a product composed of carbon, hydrogen, and oxygen. The product that was purified and sent off for elemental analysis giving the following mass percents: 68.85% C and 4.95% H. Determine the empirical formula of this compound.

In the Haber process, hydrogen gas reacts with nitrogen gas to produce ammonia. How many kilograms of hydrogen would be required to react completely with 1.0 kg of nitrogen, and how many kilograms of ammonia would be formed

What is the theoretical yield of PI3 from the reaction of 27.0 g of P and 68.0 g of I2 2P(s) + 3I2(s) →\rarr→ 2PI3(s)

When a 0.952 g sample of an organic compound containing C, H, and O is burned completely in oxygen, 1.35 g of CO2 and 0.826 g of H2O are produced. What is the empirical formula of the compound

Acetylene gas, HCCH(g) , can be generated in the laboratory by adding calcium carbide to excess water, as shown in the following reaction CaC2(s) + H2O(l) →\rarr→ HCCH(g) + CaO(s) How many grams of CaC2 would be required to generate 0.20 moles of HCCH(g)

The percent composition by mass of tartaric acid is: 32.01% C, 4.03% H, and 63.96% O. Given that the molecular mass of tartaric acid is 150 amu, determine its molecular formula.

Ammonium nitrate decomposes to give dinitrogen monoxide and water as shown in the following reaction: NH₄NO₃ $\rarr$ N₂O + 2H₂O If a 108 g sample of NH₄NO₃ decomposes to give 23 g of N₂O(g) , what percent of the original sample remains

Determine the number of moles of water produced by the reaction of 155 g of ammonia and 356 g of oxygen.4NH₃ + 5O₂ $\rarr$ 4NO + 6H₂O

When a 0.860 g sample of an organic compound containing C, H, and O was burned completely in oxygen, 1.64 g of CO₂ and 1.01 g of H₂O were produced. What is the empirical formula of the compound

What is the theoretical yield of PI₃ from the reaction of 27.0 g of P and 68.0 g of I₂ 2P(s) + 3I₂(s) $\rarr$ 2PI₃(s)

When a 0.952 g sample of an organic compound containing C, H, and O is burned completely in oxygen, 1.35 g of CO₂ and 0.826 g of H₂O are produced. What is the empirical formula of the compound

Acetylene gas, HCCH(g) , can be generated in the laboratory by adding calcium carbide to excess water, as shown in the following reaction CaC₂(s) + H₂O(l) $\rarr$ HCCH(g) + CaO(s) How many grams of CaC₂ would be required to generate 0.20 moles of HCCH(g)