Practice Questions

Gas volumes at the same conditions follow mole ratios. Three volumes of H₂ produce two volumes of NH₃. Therefore, 67.2 L H₂ (3 × 22.4 L) yields 44.8 L NH₃ (2 × 22.4 L). Concept tested: Gas volume stoichiometry at STP.

Balanced equation coefficients represent mole ratios and serve as conversion factors in stoichiometric calculations. Concept tested: Stoichiometric conversion factors.

Carbon has a molar mass of 12 g mol⁻¹. Thus, 24 g = 2 moles of C, producing 2 moles (44.8 L) of CO₂ at STP. Concept tested: Integrated mass-mole-volume calculation.

Since one mole occupies 22.4 L at STP, 11.2 L corresponds to 0.50 mole. Concept tested: Volume-to-mole conversion.

The balanced equation shows that 2 moles of H₂O₂ produce 1 mole of O₂. Hence, 6 moles produce 3 moles of oxygen gas. Concept tested: Stoichiometric relationships.

According to the balanced equation, 2 moles of Mg react with 1 mole of O₂. Therefore, 4 moles of Mg require 2 moles of O₂. Concept tested: Mole ratio application.

At STP, one mole of any ideal gas occupies 22.4 L. Thus, 2.5 × 22.4 = 56.0 L. Concept tested: Molar volume at STP.

One mole contains 6.02 × 10²³ molecules. Therefore, 0.5 mole contains half of this number, 3.01 × 10²³ molecules. Concept tested: Avogadro's number.

The balanced equation indicates a 1 : 1 mole ratio between CaCO₃ and CO₂. Therefore, 2.5 moles of CaCO₃ yield 2.5 moles of CO₂. Concept tested: Stoichiometric mole calculations.

The balanced equation shows a 2 : 2 ratio between Na and NaCl, which simplifies to 1 : 1. Therefore, 5 moles of Na produce 5 moles of NaCl. Concept tested: Mole ratio from a balanced equation.