Answer:
[tex]\large \boxed{\text{0.2797 mol}}[/tex]
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[tex]\text{Moles of CaCO}_{3} = \text{27.9 g CaCO}_{3}\times \dfrac{\text{1 mol CaCO}_{3}}{\text{100.09 g CaCO}_{3}}= \text{0.2787 mol CaCO}_{3}[/tex]
(b) Moles of CaCl₂[tex]\text{Moles of CaCl}_{2} =\text{0.2787 mol CaCO}_{3} \times \dfrac{\text{1mol CaCl}_{2}}{\text{1 mol CaCO}_{3}} = \text{0.279 mol CaCl}_{2}[/tex][tex]\text{The reaction produces $\large \boxed{\textbf{0.279 mol}}$ of CaCl}_{2}[/tex]
