Answer:
The answer to your question is 29.7 g of CaCl₂
Explanation:
data
mass CaCl₂ = ?
mass CaCO₃ = 27 g
mass HCl = 12 g
Balanced Reaction
CaCO₃ + HCl ⇒ CaCl₂ + H₂O + CO₂
Molecular weight
CaCO₃ = 40 + 12 + 48 = 100 g
HCl = 36 + 1 = 37 g
- Calculate proportions
Theoretical proportion CaCO₃ / HCl = 100 / 37 = 2.7
Experimental proportion CaCO₃/ HCl = 27/12 = 2.25
From the proportions we determine that the limiting reactant is CaCO₃, because the proportion diminishes.
Molecular weight CaCl₂ = 40 + 35.5 = 110 g
100 g CaCO₃ -------------------- 110 g of CaCl₂
27 g CaCO₃ ------------------------- x
x = (27 x 110) / 100
x = 29.7 g of CaCl₂
