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? P

Question

3 years ago

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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? P

lease show all of your work by filling in the appropriate values in the spaces in the two calculations below and then typing the mathematical result after each equals sign. Finally, answer the question by typing the answer in the space provided below the two calculations.
CaCl2 (aq) + Na2CO3 (aq) → CaCO3 (s) + 2 NaCl (aq)

Chemistry

1 answer:

Andre45 [30] · Answer 1 · 2022-05-30T23:52:15+03:00

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