Question

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

Answer 1

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

Step-by-step 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