basic biochemical calculation

Question

What is the maximal volume of CO2 (at normal conditions) which can be produced if the following reagents are available?

a) 1.00 kg of 5.0% hydrochloric acid and 200 g of limestone containing 90% of CaCO3;

 b) 1.00 kg of 5.0% sulfuric acid and 200 g of limestone containing 90% of CaCO3.

Answer 1

a). CaCO₃(s) + 2HCl ==> CaCl₂(aq) + H₂O(l) + CO₂(g)

molar mass HCl = 36.5 g / mol

molar mass CaCO3 = 100 g / mol

moles HCl present = 1.00 kg x 1000 g / kg x 5% x 1 mol / 36.5 g = 1.37 mols HCl

moles CaCO₃ present = 200 g x 90% x 1 mol / 100 g = 1.80 mols CaCO₃

Limiting reactant is HCl (2:1 mol ratio), and mols of CO2 formed is determined by HCl

mols CO₂ formed = 1.37 mols HCl x 1 mol CO2 / 2 mol HCl = 0.685 mols CO₂

Assuming normal conditions to be defined as 20ºC and 1 atm, we can use the ideal gas law to find the volume of CO₂. If by normal conditions you mean standard temperature and pressure (STP), then simply use 22.4 L /mole to find the volume.

PV = nRT

V =nRT / P = (0.685 mol)(0.0821 Latm/Kmol)(293K) / 1 atm

V = 16.5 L

(b). H₂SO₄(aq) + CaCO₃(s) ==> CaSO₄(aq) + H₂O(l) + CO₂(g)

molar mass H₂SO₄ = 98.1 g / mol

molar mass CaCO₃ = 100 g / mol

mols H₂SO₄ = 1.00 kg x 1000 g / kg x 5% x 1 mol / 98.1 g = 0.510 mols H₂SO₄

moles CaCO₃ present = 200 g x 90% x 1 mol / 100 g = 1.80 mols CaCO₃

Limiting reactant is H₂SO₄ (mol ratio 1:1) and mols CO2 formed is determined by H₂SO₄

mols CO₂ formed = 0.510 mols H₂SO₄ x 1 mol CO2 / mol H₂SO₄

PV = nRT

V = nRT / P = (0.510 mols)(0.0821 Latm/Kmol)(293K) / 1 atm

V = 12.3 L