If 6.00 g of CaCl2 • 2 H2O and 5.50 g of Na2CO3 are allowed to react in aqueous solution, what mass of CaCO3 will be produced? Answer the question by typing the answer in the space provided below the two calculations.

(6.00 g CaCl2•2 H2O/1)*(1 mole CaCl2•2 H2O/ g CaCl2•2 H2O)*( mole CaCO3/ mole CaCl2•2 H2O)*( g CaCO3/1 mole CaCO3) = g CaCO3

(5.50 g Na2CO3/1)*(1 mole Na2CO3/ g Na2CO3)*( mole CaCO3/ mole Na2CO3)*( g CaCO3/1 mole CaCO3) = g CaCO3

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

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

Explanation:

Given data:

Mass of CaCl₂.2H₂O = 6.00 g

Mass of Na₂CO₃ = 5.50 g

Mass of CaCO₃ produced = ?

Solution:

Chemical equation:

CaCl₂  +  Na₂CO₃   →   CaCO₃ + 2NaCl

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

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

Now we will compare the moles of CaCO₃  with Na₂CO₃  and CaCl₂ .

                     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

                      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