How many grams of CaCl2 could be obtained if 15.0 g HCl is allowed to react with excess CaCO3? Balanced Equation: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)

Question

Basic Stoichiometry

I am really confused /

Answer 1

No reason to be confused. Follow along....

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

Because CaCO₃ is in excess, we need not worry about it. So, HCl is limiting and will determine how much product (CaCl₂ and H₂O) can be formed.

Start by finding the moles of HCl present:

15.0 g HCl x 1 mol HCl / 36.5 g = 0.4109 moles HCl

Next, convert this to moles of CaCl₂ that can be formed from 0.4109 moles of HCl:

0.4109 mol HCl x 1 mol CaCl₂ / 2 mol HCl = 0.2055 moles CaCl₂ can be formed

Finally, convert this # moles to grams CaCl₂ using molar mass CaCl₂ of 111 g/mol

0.2055 mol CaCl₂ x 111 g /mol = 22.8 g CaCl₂ can be formed (3 sig. figs.)

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If your teacher wants you to do these calculations in a single operation (as many do), here it is....

Using stoichiometry and dimensional analysis:

15.0 g HCl x 1 mol HCl/36.5 g x 1 mol CaCl₂ / 2 mol HCl x 111 g CaCl₂/mol = 22.8 g CaCl₂