Practice Questions

The mole serves as a bridge between the atomic scale and laboratory measurements involving mass, particles, and gas volume. It does not replace balanced equations or apply only to gases. Concept tested: Significance of the mole concept.

One volume of N₂ produces two volumes of NH₃ under the same conditions. Therefore, 44.8 L of N₂ yields 89.6 L of NH₃. Concept tested: Gas volume stoichiometry.

Representative particles depend on the species present. One mole of Ca²⁺ contains Avogadro's number of calcium ions, not atoms. Concept tested: Representative particles.

The excess reagent is present in greater quantity than required and remains unreacted after the limiting reagent is completely consumed. Concept tested: Limiting and excess reagents.

According to the balanced equation, 2 volumes of H₂ react with 1 volume of O₂. Therefore, 44.8 L H₂ requires 22.4 L O₂. Concept tested: Gas volume ratios at STP.

Molar mass of CaCO₃ = 100 g mol⁻¹. Thus, 200 g = 2 moles, producing 2 moles of CO₂ because the mole ratio is 1 : 1. Concept tested: Mass-to-mole conversion using balanced equations.

Two moles contain 2 × 6.02 × 10²³ = 1.20 × 10²⁴ molecules. The remaining values correspond to incorrect multiples. Concept tested: Avogadro's number.

One mole of gas occupies 22.4 L at STP. Therefore, 0.25 × 22.4 = 5.6 L. Concept tested: Molar volume of gases.

The balanced equation shows that 2 moles of CO require 1 mole of O₂. Therefore, 5 moles of CO require 2.5 moles of O₂. Concept tested: Mole ratio application.

Molar mass of Na = 23 g mol⁻¹. Therefore, 9.2 g Na = 0.40 mole. According to the balanced equation, 2 moles of Na produce 1 mole of H₂. Thus, 0.40 mole Na produces 0.20 mole H₂. Concept tested: Mass-to-mole stoichiometric calculation.