Respuesta :
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 CaClâ‚‚.2Hâ‚‚O = 6.00 g
Mass of Na₂CO₃ = 5.50 g
Mass of CaCO₃ produced = ?
Solution:
Number of moles of CaClâ‚‚.2Hâ‚‚O.
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
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