Question

You conduct an experiment and collect 21.0g of CO₂ that was produced during the following (balanced) reaction:

2HCl + CaCO₃ ---> CO₂ + H₂O + CaCl₂
to the nearest tenth of a gram, how much calcium carbonate was used in the reaction?

Answer 1

The mass of calcium carbonate used in the reaction is 47.7 grams when 21.0 grams of CO2 are collected. The calculation is based on stoichiometry and using the molar masses of CO2 and CaCO3.

Step-by-step explanation:

To calculate the mass of calcium carbonate used, we can apply the stoichiometry of the balanced chemical equation. The molar mass of CO2 (44.01 g/mol) and CaCO3 (100.09 g/mol) are needed to perform this calculation.

Firstly, determine the moles of CO2 produced using its molar mass:

21.0 g CO2 × (1 mol CO2 / 44.01 g CO2) = 0.477 mol CO2.

According to the reaction stoichiometry, 1 mole of CaCO3 produces 1 mole of CO2, so 0.477 mol of CO2 would come from 0.477 mol of CaCO3.

Thus, the mass of CaCO3 used is calculated by:

0.477 mol CaCO3 × (100.09 g CaCO3 / 1 mol CaCO3) = 47.7 g of CaCO3 to the nearest tenth of a gram.