Answer:
6.00 g CaCl₂ .2H₂O /1 × 1 mol CaCl₂ .2H₂O / 147 g CaCl₂ .2H₂O × 0.04 mol CaCO₃/ 0.04 mol of CaCl₂ .2H₂O × 100 g CaCO₃ / 1 mole CaCO₃ = 4 g
5.50 g Na₂CO₃ /1 × 1 Na₂CO₃ / 106 g Na₂CO₃ × 0.05 mol CaCO₃/ 0.05 mol of Na₂CO₃ × 100 g CaCO₃ / 1 mole CaCO₃ = 5 g
Explanation:
Given data:
Mass of CaCl₂.2H₂O = 6.00 g
Mass of Na₂CO₃ = 5.50 g
Mass of CaCO₃ produced = ?
Solution:
Number of moles of CaCl₂.2H₂O.
Number of moles = mass/ molar mass
Number of moles = 6.00 g/ 147 g/ mol
Number of moles = 0.04 mol
Number of moles of Na₂CO₃:
Number of moles = mass/ molar mass
Number of moles = 5.50 g/ 106 g/ mol
Number of moles = 0.05 mol
Chemical equation:
CaCl₂ + Na₂CO₃ → CaCO₃ + 2NaCl
Now we will compare the moles of CaCO₃ with Na₂CO₃ and CaCl₂ through balanced chemical equation .
CaCl₂ : CaCO₃
1 : 1
0.04 : 0.04
Mass of CaCO₃:
Mass = number of moles × molar mass
Mass = 0.04 mol× 100 g/mol
Mass = 4 g
6.00 g CaCl₂ .2H₂O /1 × 1 mol CaCl₂ .2H₂O / 147 g CaCl₂ .2H₂O × 0.04 mol CaCO₃/ 0.04 mol of CaCl₂ .2H₂O × 100 g CaCO₃ / 1 mole CaCO₃ = 4 g
Na₂CO₃ : CaCO₃
1 : 1
0.05 : 0.05
Mass of CaCO₃:
Mass = number of moles × molar mass
Mass = 0.05 mol× 100 g/mol
Mass = 5 g
5.50 g Na₂CO₃ /1 × 1 Na₂CO₃ / 106 g Na₂CO₃ × 0.05 mol CaCO₃/ 0.05 mol of Na₂CO₃ × 100 g CaCO₃ / 1 mole CaCO₃ = 5 g
, where 
(in some books 6.022).
