Answer:
Limiting reactant = O₂
Excess reactant = NO
Theoretical yield of NO₂ = 46 g
Mass of excess reactant = 30 g
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Explanation:
O₂ + 2NO → 2NO₂
Mole ratio for the reaction is;
1 : 2 → 2
mass of O₂ = 16 g
mass of NO = 80 g
mass of NO₂ = 25 g
molecular weight of O₂ = 32 g/mol
molecular weight of NO = 30 g/mol
molecular weight of NO₂ = 46 g/mol
molar mass of O₂ = mass ÷ molecular weight = 16 g ÷ 32 g/mol = 0.5 mol
molar mass of NO = mass ÷ molecular weight = 80 g ÷ 30 g/mol = 2.67 mol
Since, 1 mole of O₂ requires 2 moles of NO for the combustion reaction, 0.5 mole shall require 1 mole of NO for the reaction. Thus, O₂ is the limiting reactant and NO is the excess reactant as it has an excess of 2.67 mol - 1 mol = 1.67 mol.
<h3>Theoretical yield of NO₂</h3><h3 />
1 mole of O₂ shall yield 2 moles of NO₂
Thus, 0.5 mole of O₂ shall yield 1 mole of NO₂
mass of NO₂ = molecular weight * molar mass = 46 g/mol * 1 mole = 46 g
<h3>Mass of Excess Reactant</h3>
1 mole of O₂ shall react with 2 moles of NO
Thus, 0.5 mole of O₂ shall yield 1 mole of NO
mass of NO = molecular weight * molar mass = 30 g/mol * 1 mole = 30 g
