Question

If 27.9 grams of calcium carbonate reacts with excess HCl, how many moles of calcium chloride are produced

Answer 1

`\large \boxed{\text{0.2797 mol}}`

Step-by-step explanation:

We must do the conversions:

mass of CaCO₃ ⟶ moles of CaCO₃ ⟶ moles of CaCl₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 100.09 110.98

CaCO₃ + 2HCl ⟶ CaCl₂ + CO₂ + 2HCl

m/g: 27.9

(a) Moles of CaCO₃
`\text{Moles of CaCO}_(3) = \text{27.9 g CaCO}_(3)* \frac{\text{1 mol CaCO}_(3)}{\text{100.09 g CaCO}_(3)}= \text{0.2787 mol CaCO}_(3)`

(b) Moles of CaCl₂
`\text{Moles of CaCl}_(2) =\text{0.2787 mol CaCO}_(3) * \frac{\text{1mol CaCl}_(2)}{\text{1 mol CaCO}_(3)} = \text{0.279 mol CaCl}_(2)`
`\text{The reaction produces $\large \boxed{\textbf{0.279 mol}}$ of CaCl}_(2)`