Question

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+H2O(l)+CO2(g) How many grams of calcium chloride will be produced when 31.0 g of calcium carbonate is combined with 13.0 g of hydrochloric acid?

Answer 1

Answer:

19.7822 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For calcium carbonate

Given mass = 31.0 g

Molar mass of calcium carbonate = 100.0869 g/mol

Moles of calcium carbonate = 31.0 g / 100.0869 g/mol = 0.3097 moles

Given: For hydrochloric acid

Given mass = 13.0 g

Molar mass of hydrochloric acid = 36.46094 g/mol

Moles of hydrochloric acid = 13.0 g / 36.46094 g/mol = 0.3565 moles

According to the given reaction:

$CaCO_3_{(s)}+2HCl_{(aq)}\rightarrow CaCl_2_{(aq)}+H_2O_{(l)}+CO_2_{(g)}$

1 mole of calcium carbonate react with 2 moles of hydrochloric acid

0.3097 moles  of calcium carbonate react with 2 × 0.3097 moles of hydrochloric acid

Moles of HCl = 0.6194 moles

Available moles of HCl = 0.3565 moles

Limiting reagent is the one which is present in small amount. Thus, HCl is limiting reagent. (0.3565 < 0.6194)

The formation of the product is governed by the limiting reagent. So,

2 moles of HCl produces 1 mole of calcium chloride

1 mole of HCl produces $\frac {1}{2}$ mole of calcium chloride

0.3565 moles  of HCl produces $\frac {1}{2}\times 0.3565$ mole of calcium chloride

Moles of calcium chloride = 0.17825 moles

Molar mass of calcium chloride = 110.98 g/mol

Mass of calcium chloride = Moles × Molar mass = 0.17825 × 110.98 g = 19.7822 g