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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.
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.
Representative particles depend on the species present. One mole of Ca²⁺ contains Avogadro's number of calcium ions, not atoms. Concept tested: Representative particles.
At STP, one mole of gas occupies 22.4 L. Therefore, 89.6 ÷ 22.4 = 4 moles. Concept tested: Gas volume to mole conversion.
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.
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 molar mass of CO₂ is 44 g mol⁻¹. Therefore, 44 g corresponds to exactly one mole. Concept tested: Mass-to-mole conversion.
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 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 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.
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 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.
One mole of gas occupies 22.4 L at STP. Therefore, 0.25 × 22.4 = 5.6 L. Concept tested: Molar volume of gases.
Two moles contain 2 × 6.02 × 10²³ = 1.20 × 10²⁴ molecules. The remaining values correspond to incorrect multiples. Concept tested: Avogadro's number.
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.
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.
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 moles of HgO produce 1 mole of O₂. Therefore, 8 moles of HgO produce 4 moles of O₂. Concept tested: Stoichiometric calculations.
Stoichiometric calculations depend on the coefficients of a balanced equation because they represent mole ratios. Physical states, temperature, and reaction rate are not sufficient for quantitative calculations. Concept tested: Importance of balanced equations in stoichiometry.
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