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?

1) CO₂

2) 0.2551 g

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

answer: c.

explanation: