Question

For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium carbonate?

Answer 1

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

• Na₂CO₃: 1 mole
• Ca(NO₃)₂: 1 mole
• CaCO₃: 1 mole
• NaNO₃: 2 mole

Being the molar mass of the compounds:

• Na₂CO₃: 106 g/mole
• Ca(NO₃)₂: 164 g/mole  
• CaCO₃: 100 g/mole
• NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

• Na₂CO₃: 1 mole* 106 g/mole= 106 g
• Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
• CaCO₃: 1 mole* 100 g/mole= 100 g
• NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of  CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of  CaCO₃?

$mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}$

mass of CaCO₃= 74.81 grams

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.