Quiz 2 - Chemical Calculations | Chemical Calculations

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.

The reaction requires 1 mole of O₂​ for every 2 moles of H₂​. With 4 moles of H₂, 2 moles of O₂ are required. Since you have exactly 2 moles of O₂​, H₂ and O₂ are introduced in stoichiometric amounts, and neither is the limiting reactant.

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%.

• The molar mass of H₂O is 2 x 1.01 + 16.00 = 18.022 g/mol.
• The mass of hydrogen in H₂O 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%.

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.

• Dilution formula: C₁V₁ = C₂V₂​, where C₁ = 2.00 M, V₁ = 50.0 mL, and V₂ = 200.0 mL
• C₂ = $\frac{{\mathrm{C}}_1{\mathrm{V}}_1}{{\mathrm{V}}_2}$ = $\frac{2.00 \mathrm{M} \mathrm{x} 50.0 \mathrm{mL}}{200.0 \mathrm{mL}}$ = 0.50 M

• The reaction between NaOH and HCl is a 1:1 mole ratio: NaOH + HCl → NaCl + H₂O
• 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​ = C₂​V₂, where C₁ = 0.100 M, V₁ = 25.0 mL, V₂ = 50.0 mL.
• C₂ = $\frac{{\mathrm{C}}_1{\mathrm{V}}_1}{{\mathrm{V}}_2}$ = $\frac{0.100 \mathrm{M} \mathrm{x} 25.0 \mathrm{mL}}{50.0 \mathrm{mL}}$ = 0.050 M​

• Molar mass of Fe₂O₃ 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 Fe₂O₃​. Therefore, 2 moles of Fe will require 1 mole of Fe₂O₃​ (159.7 g) ⇒ Hence, 159.7 x 2 = 320 g of Fe₂O₃​ is required to produce 112 g of iron.

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

The balanced chemical equation for the neutralization reaction between sulfuric acid (H₂SO₄​) and sodium hydroxide (NaOH) is:
H_2​SO₄​ + 2 NaOH → Na₂​SO₄​ + 2 H₂​O
From the balanced equation, we see that 1 mole of H₂SO₄​ reacts with 2 moles of NaOH. Therefore, the molar ratio of H₂SO₄​ to NaOH is 1:2. The given amounts, 0.1 moles of H₂SO₄ and 0.2 moles of NaOH, confirm this ratio.