How much Calcium oxide is formed when 82 g of calcium nitrate is heated? Also find the volume of nitrogen dioxide evolved: 2Ca(NO₃)₂ ⟶ 2CaO + 4NO₂ + O₂ [Ca = 40, N = 14, O = 16]
Topic: Mole concept, gas volumes and stoichiometry
Answer
2Ca(NO₃)₂ ⟶ 2CaO + 4NO₂ + O₂
Molar mass of Ca(NO₃)₂ = 40 + 2(14 + 48) = 40 + 124 = 164 g mol⁻¹ Molar mass of CaO = 40 + 16 = 56 g mol⁻¹
Moles of Ca(NO₃)₂ = 82 ÷ 164 = 0.5 mol
From the equation, 2 mol Ca(NO₃)₂ give 2 mol CaO and 4 mol NO₂.
Mass of CaO: moles of CaO = 0.5 mol (same ratio, 2 : 2) mass = 0.5 × 56 = 28 g
Volume of NO₂ at STP: moles of NO₂ = 0.5 × (4 ÷ 2) = 1 mol volume = 1 × 22.4 = 22.4 litres
28 g of calcium oxide and 22.4 litres of nitrogen dioxide at STP.
Work in moles, never in grams directly. Grams convert to moles, moles travel across the equation using the coefficients, and then moles convert back to whatever the question asks for — grams for a solid, litres for a gas.
Two traps here. First, the coefficients are 2 and 4, not 1 and 1: the NO₂ is FOUR times the 2 mol of nitrate, which is twice the 0.5 mol you started with. Miss that and you get 11.2 litres. Second, do not compute the mass of NO₂ and then try to turn it into a volume — go straight from moles to 22.4 litres per mole.