# Quiz 2 - Chemical Calculations | Chemical Calculations

## General Chemistry 2 - Quiz 2 - Chemical Calculations

What is the mass of 3.00 moles of NaCl?

The molar mass of NaCl is 58.5 g/mol. Therefore, the mass of 3.00 moles of NaCl is 3.00 x 58.5 = 175.5 g.

In the reaction 2 H_{2} + O_{2} → 2 H_{2}O, if you start with 4 moles of H_{2} and 2 moles of O_{2}, what is the limiting reactant?

The reaction requires 1 mole of O_{2} for every 2 moles of H_{2}. With 4 moles of H_{2}, 2 moles of O_{2} are required. Since you have exactly 2 moles of O_{2}, H_{2} and O_{2} are introduced in stoichiometric amounts, and neither is the limiting reactant.

If the theoretical yield of a reaction is 10.0 g and the actual yield is 8.0 g, what is the percent yield?

Percent yield is calculated as: $\frac{\mathrm{actual}\mathrm{yield}}{\mathrm{theoretical}\mathrm{yield}}$ x 100.

Therefore, $\frac{8.0\mathrm{g}}{10.0\mathrm{g}}$ x 100 = 80%.

What is the percent composition by mass of hydrogen in water?

- The molar mass of H
_{2}O is 2 x 1.01 + 16.00 = 18.022 g/mol. - The mass of hydrogen in H
_{2}O is 2 x 1.01 = 2.022 g/mol. - Therefore, the percent composition by mass of hydrogen is $\frac{2.02}{18.02}$ x 100 = 11.1%.

Which of the following is true about molarity (M) and molality (m)?

Molarity (*M*) is defined as the number of moles of solute per liter of solution, while molality (*m*) is defined as the number of moles of solute per kilogram of solvent.

If 50.0 mL of 2.00 M HCl is diluted to a final volume of 200.0 mL, what is the final concentration?

- Dilution formula: C
_{1}V_{1 }= C_{2}V_{2}, where C_{1}= 2.00 M, V_{1 }= 50.0 mL, and V_{2 }= 200.0 mL - C
_{2}= $\frac{{\mathrm{C}}_{1}{\mathrm{V}}_{1}}{{\mathrm{V}}_{2}}$ = $\frac{2.00\mathrm{M}\mathrm{x}50.0\mathrm{mL}}{200.0\mathrm{mL}}$ = 0.50 M

In an acid-base titration, 25.0 mL of 0.100 M NaOH is required to neutralize 50.0 mL of HCl. What is the concentration of the HCl solution?

- The reaction between NaOH and HCl is a 1:1 mole ratio: NaOH + HCl → NaCl + H
_{2}O - At the endpoint of the titration, the number of moles of acid is equal to the number of moles of base: C
_{1}V_{1 }= C_{2}V_{2}, where C_{1}= 0.100 M, V_{1}= 25.0 mL, V_{2}= 50.0 mL. - C
_{2}= $\frac{{\mathrm{C}}_{1}{\mathrm{V}}_{1}}{{\mathrm{V}}_{2}}$ = $\frac{0.100\mathrm{M}\mathrm{x}25.0\mathrm{mL}}{50.0\mathrm{mL}}$ = 0.050 M

Given the balanced equation 2 Fe_{2}O_{3} + 3 C → 4 Fe + 3 CO_{2}, how many grams of Fe_{2}O_{3} are required to produce 112 grams of iron? (Atomic masses: Fe = 55.85, O = 16.00)

- Molar mass of Fe
_{2}O_{3}is 2 x 55.85 + 3 x 16.00 = 159.7 g/mol. - Moles of 112 grams of Fe: n
_{Fe}= $\frac{{\mathrm{m}}_{\mathrm{Fe}}}{{\mathrm{M}}_{\mathrm{Fe}}}$ = $\frac{112\mathrm{g}}{55.85\mathrm{g}.{\mathrm{mol}}^{-1}}$ = 2 mol - From the equation, 4 moles of Fe are produced from 2 moles of Fe
_{2}O_{3}. Therefore, 2 moles of Fe will require 1 mole of Fe_{2}O_{3} (159.7 g) ⇒ Hence, 159.7 x 2 = 320 g of Fe_{2}O_{3} is required to produce 112 g of iron.

A solution contains 0.5 moles of NaCl dissolved in 250 mL of water. What is the molarity of the solution?

Molarity (M) is defined as moles of solute per liter of solution. Therefore, M = $\frac{\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{L}}$ = $\frac{0.5}{0.250}$ = 2 M

In a titration, 0.1 moles of H_{2}SO_{4} is required to neutralize 0.2 moles of NaOH. What is the molar ratio of H_{2}SO_{4} to NaOH in this reaction?

The balanced chemical equation for the neutralization reaction between sulfuric acid (H_{2}SO_{4}) and sodium hydroxide (NaOH) is:

H_{2}SO_{4} + 2 NaOH → Na_{2}SO_{4} + 2 H_{2}O

From the balanced equation, we see that 1 mole of H_{2}SO_{4} reacts with 2 moles of NaOH. Therefore, the molar ratio of H_{2}SO_{4} to NaOH is 1:2. The given amounts, 0.1 moles of H_{2}SO_{4} and 0.2 moles of NaOH, confirm this ratio.