Consider the reaction of CaC2 and water to produce CaCO3 and NH3 according to the reaction CaCN2 + 3H2O → CaCO3 + 2 NH3. How much CaCO3 is produced upon reaction of 45 g CaCN2 and 45 g of H2O

16.27 g of CaCO3 are produced upon reaction of 45 g CaCN2 and 45 g of H2O.

Explanation:

Ca (CN) 2 + 3H2O → CaCO3 + 2 NH3

First of all, let's find out the limiting reactant.

Molar mass Ca (CN) 2.

Molar mass H2O: 18 g/m

Moles of Ca (CN) 2: mass / molar mass

45 g / 92.08 g/m = 0.488 moles

Moles of H2O: mass / molar mass

45g / 18g/m = 2.50 moles

This is my rule of three

1 mol of Ca (CN) 2 needs 3 moles of H2O

2.5 moles of Ca (CN) 2 needs (2.5. 3) / 1 = 7.5 moles

I need 7.5 moles of water, but I only have 0.488. Obviously water is the limiting reactant; now we can work on it.

3 moles of water __ makes __ 1 mol of CaCO3

0.488 moles of water __ makes ___ (0.488. 1) / 3 = 0.163 moles

Molar mass CaCO3 = 100.08 g/m

Molar mass. moles = mass

100.08 g/m. 0.163 moles = 16.27 g