Question

A mix of 0.5236 gr qe contains 42.30% calcium carbonate and 57.70% magnesium carbonate. This sample was treated with hydrochloric acid and the gas produced was collected.

1) what is the gas collected
2) how many grams of gas were produced?

Answer 1

1) CO₂

2) 0.2551 g

Step-by-step explanation:

The balanced reactions are:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

MgCO₃ + 2HCl → MgCl₂ + H₂O + CO₂

1) The gas produced is CO₂.

2) Calculate mass of CaCO₃:

(0.5236 g) (0.4230) = 0.2215 g CaCO₃

Convert to moles:

(0.2215 g CaCO₃) (1 mol / 100.1 g) = 0.002213 mol CaCO₃

Find moles of CaCO₃:

(0.002213 mol CaCO₃) (1 mol CO₂ / mol CaCO₃) = 0.002213 mol CO₂

Convert to mass:

(0.002213 mol CO₂) (44.01 g / mol) = 0.09738 g CO₂

Calculate mass of MgCO₃:

(0.5236 g) (0.5770) = 0.3021 g MgCO₃

Convert to moles:

(0.3021 g MgCO₃) (1 mol / 84.31 g) = 0.003583 mol MgCO₃

Find moles of MgCO₃:

(0.003583 mol MgCO₃) (1 mol CO₂ / mol MgCO₃) = 0.003583 mol CO₂

Convert to mass:

(0.003583 mol CO₂) (44.01 g / mol) = 0.1577 g CO₂

Total mass of CO₂:

0.09738 g CO₂ + 0.1577 g CO₂ = 0.2551 g CO₂