MCQs

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.

The coefficients in a balanced chemical equation represent the relative numbers of molecules or moles participating in the reaction. Masses depend on molar masses and are not directly represented by the coefficients. Concept tested: Significance of mole ratios in stoichiometry.

The balanced equation shows that 1 mole of Ca reacts with 2 moles of HCl. Therefore, 3 moles of Ca require 6 moles of HCl. Concept tested: Mole ratio application.

The limiting reagent is the reactant that is consumed first and determines the maximum amount of product formed. It is not necessarily the reactant with the smallest mass or highest molar mass. Concept tested: Limiting reagent concept.

The molar mass of CO₂ is 44 g mol⁻¹. Therefore, 44 g corresponds to exactly one mole. Concept tested: Mass-to-mole conversion.

One mole contains 6.02 × 10²³ molecules. Therefore, 0.25 mole contains one-fourth of this value, equal to 1.51 × 10²³ molecules. Concept tested: Representative particles.

The equation shows that 2 moles of K produce 1 mole of H₂. Therefore, 8 moles of K produce 4 moles of hydrogen gas. Concept tested: Stoichiometric mole ratio.

At STP, one mole of gas occupies 22.4 L. Therefore, 89.6 ÷ 22.4 = 4 moles. Concept tested: Gas volume to mole conversion.

According to the balanced equation, 2 moles of HgO produce 1 mole of O₂. Therefore, 8 moles of HgO produce 4 moles of O₂. Concept tested: Stoichiometric calculations.

Three moles contain 3 × 6.02 × 10²³ = 1.81 × 10²⁴ molecules. The other values represent fewer moles. Concept tested: Mole-to-particles conversion.

The balanced equation shows that 3 moles of Cl₂ produce 2 moles of AlCl₃. Therefore, 6 moles of Cl₂ produce 4 moles of AlCl₃. Concept tested: Mole ratio from a balanced equation.

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.