16.0 grams of oxygen gas reacted with 80.0 grams nitrogen monoxide gas producing 25.0 grams of nitrogen dioxide gas in the lab. Determine the limiting and excess reactant, calculate the theoretical yield, calculate the number of grams of excess reactant

Limiting reactant  = O₂

Excess reactant = NO

Theoretical yield of NO₂ = 46 g

Mass of excess reactant = 30 g

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.

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

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

pv=nrt

explanation: