Question 1

What is the percentage of calcium in an ionic compound if a 1.23 g sample completely dissolves in water and yields 0.893 g of calcium carbonate upon addition of excess sodium carbonate solution?

Accepted Answer

First, we need to write the balanced chemical equation for what is happening when the sample dissolves in water and then reacts with excess sodium carbonate:

CaX + Na2CO3 → CaCO3 + 2NaX

where CaX represents the calcium-containing compound that we are trying to analyze.

We know that the mass of CaCO3 produced is 0.893 g, which means that we can use stoichiometry to determine the amount (in moles) of CaX that was originally in the sample. The molar mass of CaCO3 is 100.09 g/mol, so:

0.893 g CaCO3 × (1 mol CaCO3/100.09 g CaCO3) × (1 mol CaX/1 mol CaCO3) = 0.008923 mol CaX

This represents the amount (in moles) of CaX that was originally in the 1.23 g sample. To convert this to a percentage, we need to know the molar mass of CaX. Let's call this M. Then:

1.23 g CaX × (1 mol CaX/M g CaX) = 0.008923 mol CaX

Solving for M:

M = 138.24 g/mol

So, the percentage of calcium in the compound is:

(1 mol Ca/1 mol CaX) × (40.08 g Ca/mol Ca) ÷ (138.24 g/mol CaX) × 100% = 29.1%

Therefore, answer B) is the best choice.

Question 2

Exhibit 4-3 Consider the Acid-Base neutralization of sulfuric acid with sodium hydroxide to answer the following question(s). NaOH (aq) + HS1U1B12S1U1B0SOS1U1B14S1U1B0→ -Refer to Exhibit 4-3. If 26.87 mL of a standard 0.1650 M NaOH solution is required to neutralize 39.70 mL of a H₂SO₄ solution, what is the molarity of the H₂SO₄ acid solution?

Accepted Answer

The correct answer is A.To determine the molarity of the H2SO4 solution, we can use the formula:M1V1 = M2V2where M1 and V1 represent the molarity and volume of the NaOH solution, and M2 and V2 represent the molarity and volume of the H2SO4 solution.Substituting the given values into the formula:0.1650 M * 26.87 mL = M2 * 39.70 mLSolving for M2:M2 = (0.1650 M * 26.87 mL) / 39.70 mL = 0.05584 MTherefore, the molarity of the H2SO4 acid solution is 0.05584 M.

Question 3

What is the molarity of a HCl solution if the addition of excess AgNO₃ to a 20.0 mL sample of the solution yields 0.113 g of AgCl?

Accepted Answer

The balanced chemical equation for the reaction between HCl and AgNO3 is:

HCl + AgNO3 → AgCl + HNO3

From the equation, we can see that 1 mole of HCl reacts with 1 mole of AgNO3 to produce 1 mole of AgCl. The molar mass of AgCl is 143.32 g/mol. Therefore, the number of moles of AgCl produced in the reaction is:

0.113 g AgCl × (1 mol AgCl/143.32 g AgCl) = 0.000789 mol AgCl

Since the reaction is between HCl and AgNO3 in a 1:1 mole ratio, the number of moles of HCl in the 20.0 mL sample can be calculated as follows:

0.000789 mol AgCl × (1 mol HCl/1 mol AgCl) = 0.000789 mol HCl

The volume of the sample is 20.0 mL or 0.0200 L. Therefore, the molarity of the HCl solution is:

Molarity = moles/volume = 0.000789 mol/0.0200 L = 0.0394 M

The correct answer is C) 0.0394 molar.

Question 4

What is the percent of chlorine in an ionic compound if a 1.23 g sample dissolves in water and reacts with excess AgNO₃ solution to yield 1.45 g of AgCl?

Accepted Answer

To find the percent of chlorine, first calculate the moles of AgCl formed: 1.45 g AgCl / 143.32 g/mol = 0.0101 mol. Since the ratio of AgCl to Cl is 1:1, there are 0.0101 mol of Cl. The mass of Cl is 0.0101 mol * 35.45 g/mol = 0.358 g. The percent of chlorine is (0.358 g / 1.23 g) * 100% = 29.1%, which rounds to 29.2%.

Question 5

What is the molarity of a NaOH solution, if 25.0 mL of this NaOH solution is required to neutralize 20.0 mL of 0.300 M H₂SO₄? (Write a balanced reaction to ascertain the stoichiometry of this reaction.)

Accepted Answer

The balanced reaction between NaOH and H₂SO₄ is: NaOH + HH₂SO₄ → NaH₂SO₄ + H2O From this reaction, we can see that 1 mole of NaOH reacts with 1 mole of H₂SO₄. To find the molarity of the NaOH solution, we need to use the formula: Molarity (M) = moles of solute / volume of solution (in liters) First, we need to calculate the number of moles of H₂SO₄: moles of H₂SO₄ = Molarity x volume (in liters) moles of H₂SO₄ = 0.300 M x 0.0200 L moles of H₂SO₄ = 0.0060 moles Since the reaction is 1:1, the number of moles of NaOH required to neutralize the H₂SO₄ is also 0.0060 moles. Now, we can use the formula to find the molarity of the NaOH solution: Molarity (M) = moles of NaOH / volume of NaOH solution (in liters) We know that 25.0 mL of the NaOH solution was required to neutralize the H₂SO₄. We need to convert this to liters: volume of NaOH solution = 25.0 mL = 0.0250 L Now we can plug in the values: Molarity (M) = 0.0060 moles / 0.0250 L Molarity (M) = 0.240 M Therefore, the molarity of the NaOH solution is 0.240 M, and the best choice is answer choice C.

Question 6

If 16.87 mL of 0.165 M NaOH solution is titrated witH₄9.70 mL of H₂ SO₄. What is the molar concentration of the H₂ SO₄solution? 2 NaOH (aq) + H₂ SO₄(aq) Na₂ SO₄(aq) + 2 H₂ O ( )

Accepted Answer

The answer of If 16.87 mL of 0.165 M NaOH...

Question 7

What is the molarity of a NaCl solution if the addition of excess AgNO₃ to a 40 mL sample of the solution yields 0.123 g of AgCl?

Accepted Answer

Firstly, write the balanced equation for the reaction: NaCl + AgNO3 → AgCl + NaNO3
Next, calculate the number of moles of AgCl formed: 
Mass of AgCl = 0.123 g 
Molar mass of AgCl = 143.32 g/mol 
Number of moles of AgCl = 0.123 g ÷ 143.32 g/mol = 8.575 x 10^-4 mol
Since AgNO3 is in excess, all the NaCl in the solution would be used up. Thus, the number of moles of NaCl in the sample is also 8.575 x 10^-4 mol. 
Finally, calculate the molarity of the NaCl solution: 
Molarity = number of moles ÷ volume in liters 
Volume = 40 mL ÷ 1000 = 0.04 L 
Molarity = 8.575 x 10^-4 mol ÷ 0.04 L = 0.021 molar.

Quiz 4: Chemical Reactions in Solution

What is the percentage of calcium in an ionic compound if a 1.23 g sample completely dissolves in water and yields 0.893 g of calcium carbonate upon addition of excess sodium carbonate solution?

What is the molarity of a HCl solution if the addition of excess AgNO3 to a 20.0 mL sample of the solution yields 0.113 g of AgCl?

What is the percent of chlorine in an ionic compound if a 1.23 g sample dissolves in water and reacts with excess AgNO3 solution to yield 1.45 g of AgCl?

What is the molarity of a NaOH solution, if 25.0 mL of this NaOH solution is required to neutralize 20.0 mL of 0.300 M H2SO4? (Write a balanced reaction to ascertain the stoichiometry of this reaction.)

If 16.87 mL of 0.165 M NaOH solution is titrated witH49.70 mL of H₂ SO₄. What is the molar concentration of the H₂ SO₄solution? 2 NaOH (aq) + H₂ SO₄(aq) Na2 SO₄(aq) + 2 H₂ O ( )

What is the molarity of a NaCl solution if the addition of excess AgNO3 to a 40 mL sample of the solution yields 0.123 g of AgCl?

What is the molarity of a HCl solution if the addition of excess AgNO₃ to a 20.0 mL sample of the solution yields 0.113 g of AgCl?

What is the percent of chlorine in an ionic compound if a 1.23 g sample dissolves in water and reacts with excess AgNO₃ solution to yield 1.45 g of AgCl?

What is the molarity of a NaOH solution, if 25.0 mL of this NaOH solution is required to neutralize 20.0 mL of 0.300 M H₂SO₄? (Write a balanced reaction to ascertain the stoichiometry of this reaction.)

If 16.87 mL of 0.165 M NaOH solution is titrated witH₄9.70 mL of H₂ SO₄. What is the molar concentration of the H₂ SO₄solution? 2 NaOH (aq) + H₂ SO₄(aq) Na₂ SO₄(aq) + 2 H₂ O ( )

What is the molarity of a NaCl solution if the addition of excess AgNO₃ to a 40 mL sample of the solution yields 0.123 g of AgCl?