Question 1

If 0.1800 g of impure soda ash (Na₂CO₃) is titrated with 15.66 mL of 0.1082 M HCl, what is the percent purity of the soda ash? Na₂CO₃(aq) + 2 HCl(aq) ? 2 NaCl(aq) + H₂O( $\ell$ ) + CO₂(g)

Accepted Answer

The answer of If 0.1800 g of impure soda ash...

Question 2

An impure sample of benzoic acid (C₆H₅COOH, 122.12 g/mol) is titrated with 0.5360 M NaOH. A 1.888-g sample requires 20.50 mL of titrant to reach the endpoint. What is the percent by mass of benzoic acid in the sample? C₆H₅COOH(aq) + NaOH(aq) → NaC₆H₅COO(aq) + H₂O(l)

Accepted Answer

The answer of An impure sample of benzoic acid (C₆H₅COOH,...

Question 3

The numbers preceding the formulas in chemical equations are referred to as the _____ coefficients.

Accepted Answer

The answer of The numbers preceding the formulas in chemical...

Question 4

Molarity is defined as the _____.

Accepted Answer

Molarity is defined as the amount of solute (in moles) per liter of solution.

Question 5

A 30.00-mL sample of a weak monoprotic acid is titrated with 0.0915 M NaOH. At the endpoint, it is found that 45.98 mL of titrant was used. What was the concentration of the weak acid?

Accepted Answer

First, calculate the number of moles of NaOH used in the titration:

45.98 mL × 0.0915 mol/L = 0.00421 mol NaOH

Since the acid is monoprotic, the number of moles of acid present in the solution is equal to the number of moles of NaOH used in the titration.

Therefore, the concentration of the weak acid is:

0.00421 mol / 0.03000 L = 0.140 M

So, the best choice is A.

Question 6

The following equation: A = ε × $\ell$ × C (where $\ell$ is path length and C is concentration) is known as the _____-Lambert law.

Accepted Answer

The answer of The following equation: A = ε ×...

Question 7

Which of the following is true about the actual yield of a product?

Accepted Answer

The actual yield of a product is almost always less than the theoretical yield due to factors such as incomplete reactions, side reactions, and losses during isolation/purification processes.

Question 8

A mass of 0.4113 g of an unknown acid, HA, is titrated with sodium hydroxide, NaOH. If the acid reacts with 28.10 mL of 0.1055 M aqueous sodium hydroxide, what is the molar mass of the acid?

Accepted Answer

First, we need to calculate the moles of NaOH used in the titration:

moles NaOH = volume x concentration = 0.0281 L x 0.1055 mol/L = 0.00296955 mol NaOH

Since the acid and the base react in a 1:1 ratio, the number of moles of acid is also 0.00296955 mol.

From the mass of acid given in the problem (0.4113 g), we can calculate its molar mass:

molar mass = mass / moles = 0.4113 g / 0.00296955 mol = 138.7 g/mol

Therefore, the answer is C.

Question 9

The percent yield of a chemical reaction is calculated by dividing the actual yield by the ________ yield and multiplying by 100%.

Accepted Answer

The answer of The percent yield of a chemical reaction...

Question 10

In combustion analysis, the gases produced by the combustion of a hydrocarbon are passed through a tube containing finely divided sodium hydroxide supported on asbestos. The purpose of the sodium hydroxide is to absorb ________ produced by the combustion reaction.

Accepted Answer

The answer of In combustion analysis, the gases produced by...

Question 11

The pH of purified water (or of a neutral solution) is _____.

Accepted Answer

The answer of The pH of purified water (or of...

Question 12

Which of the following is the correct representation for absorbance of a sample?

Accepted Answer

The correct formula for absorbance is A = -log T, where T is the transmittance of the sample. This formula is widely accepted and commonly used in the field of spectroscopy.

Question 13

Phosphorus trichloride, PCl₃, can be produced from the reaction below: P₄(s) + 6 Cl₂(g) → 4 PCl₃(g) If 1.00 g of phosphorus, P₄, reacts with 1.00 g of chlorine, Cl₂, which reactant is the limiting reactant and what mass of product may be produced?

Accepted Answer

The answer of Phosphorus trichloride, PCl₃, can be produced from...

Question 14

An unknown diprotic acid (H₂A) requires 31.81 mL of 0.109 M NaOH to completely neutralize a 0.685 g sample. Calculate the approximate molar mass of the acid.

Accepted Answer

First, we need to determine the moles of NaOH used:
0.03181 L x 0.109 mol/L = 0.00347 mol NaOH

Since the acid is diprotic, it will require two moles of NaOH to neutralize one mole of the acid. Therefore, the moles of acid can be calculated as:

0.00347 mol NaOH x (1 mol acid / 2 mol NaOH) = 0.00174 mol acid

To find the molar mass of the acid, we can use the formula:

Molar mass = mass / moles

The mass of the acid is given as 0.685 g, so:

Molar mass = 0.685 g / 0.00174 mol = 394.8 g/mol

Rounding to the nearest whole number gives us an approximate molar mass of 395 g/mol, which corresponds to choice B.

Question 15

A solution of sodium oxalate (Na₂C₂O₄) in acidic solution is titrated with a solution of potassium permanganate (KMnO₄) according to the following balanced chemical equation: 2KMnO₄(aq) + 8H₂SO₄(aq) + 5Na₂C₂O₄(aq) → 2MnSO₄(aq) + 8H₂O(l) + 10CO₂(g) + 5Na₂SO₄(aq) + K₂SO₄(aq) What volume of 0.0206 M KMnO₄ is required to titrate 0.176 g of Na₂C₂O₄ dissolved in 50.0 mL of solution?

Accepted Answer

First, we need to calculate the number of moles of Na2C2O4:
moles Na2C2O4 = 0.176 g / (2 * 23.00 + 12.01 + 4 * 16.00) = 0.001 mol

Next, we can use stoichiometry to determine how many moles of KMnO4 are required to react with the Na2C2O4:
2 KMnO4 + 5 Na2C2O4 + 8 H2SO4 → 2 MnSO4 + 10 CO2 + 8 H2O + 5 Na2SO4 + K2SO4

From the balanced equation, we can see that 2 moles of KMnO4 react with 5 moles of Na2C2O4. Therefore, the number of moles of KMnO4 needed is:

moles KMnO4 = (0.001 mol Na2C2O4) / (5 mol Na2C2O4 / 2 mol KMnO4) = 0.0004 mol KMnO4

Finally, we can use the molarity and volume of the KMnO4 solution to determine the volume required for the titration:

moles KMnO4 = M KMnO4 x V KMnO4
V KMnO4 = moles KMnO4 / M KMnO4 = 0.0004 mol / 0.0206 M = 0.0194 L = 19.4 mL

Therefore, the best choice is C) 25.5 mL, which is the closest to the calculated volume of KMnO4 required for the titration.

Question 16

The reaction of 5.07 g N₂ with 0.722 g H₂ produces 1.27 g NH₃. The percent yield of this reaction is ________.

Accepted Answer

The answer of The reaction of 5.07 g N₂ with...

Question 17

The reaction of HCl with NaOH is represented by the equation HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) What volume of 0.614 M HCl is required to titrate 18.8 mL of 0.619 M NaOH?

Accepted Answer

First, we need to balance the equation:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Using the equation, we can see that one mole of HCl reacts with one mole of NaOH to produce one mole of NaCl and one mole of water. Therefore, we can use the following equation to find the moles of HCl:
moles of HCl = (0.619 M NaOH) x (0.0188 L NaOH) = 0.01163 moles NaOH

Since the mole ratio of NaOH to HCl is 1:1, we know that we need 0.01163 moles of HCl to react with the NaOH. We can use the following equation to find the volume of 0.614 M HCl required to provide this many moles of HCl:
moles of HCl = (0.614 M HCl) x (volume of HCl in L)

Rearranging the equation, we can solve for the volume of HCl:
volume of HCl = moles of HCl / (0.614 M HCl) = 0.01163 moles / (0.614 mol/L) = 0.0189 L = 18.9 mL

Therefore, the closest answer choice is C) 19.0 mL.

Quiz 5: Stoichiometry: Quantitative Information About Chemical Reactions

If 0.1800 g of impure soda ash (Na2CO3) is titrated with 15.66 mL of 0.1082 M HCl, what is the percent purity of the soda ash? Na2CO3(aq) + 2 HCl(aq) ? 2 NaCl(aq) + H2O( ℓ\ellℓ ) + CO2(g)

An impure sample of benzoic acid (C6H5COOH, 122.12 g/mol) is titrated with 0.5360 M NaOH. A 1.888-g sample requires 20.50 mL of titrant to reach the endpoint. What is the percent by mass of benzoic acid in the sample? C6H5COOH(aq) + NaOH(aq) → NaC6H5COO(aq) + H2O(l)