Answer:
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
