CaCN2 + 3H2O --> CaCO3 + 2NH3?
Consider the following
CaCN2 + 3H2O --> CaCO3 + 2NH3
109.0 g CaCN2 and 82.0 g H2O are reacted. Assuming a 80.5 percent yield, which reactant is in excess and how much is leftover? The molar mass of CaCN2 is 80.11 g/mol. The molar mass of CaCO3 is 100.09 g/mol.
1. H2O; 8.46 g left over
2. CaCN2; 3.19 g left over
3. CaCN2; 0.47 g left over
4. H2O; 6.81 g left over
5. H2O; 65.9 g left over
6. CaCN2; 57.5 g left over
Comments
CaCN2 + 3 H2O → CaCO3 + 2 NH3
(109.0 g CaCN2) / (80.11 g CaCN2/mol) = 1.36063 moles CaCN2
(82.0 g H2O) / (18.0153 g H2O/mol) = 4.55169 moles H2O
1.36063 moles of CaCN2 would react completely with 3 x 1.36063 = 4.08189 moles of H2O, but there is more H2O present than that, so H2O is in excess. If the reaction proceeds only 80.5% of the way to completion, then only 4.08189 x 0.805 = 3.285921 moles of H2O were reacted, leaving 4.55169 - 3.285921 = 1.265769 moles of H2O unreacted.
(1.265769 mol H2O) x (18.0153 g/mol) = 22.8 g H2O left over
So none of the given choices apply.
However, supposing the the 80.5% yield was caused by mishandling of the product after the reaction went to actual completion, then the expression for the mass of H2O left over would be:
(4.55169 - 4.08189) x 18.0153 = 8.46 g H2O
corresponding to answer 1.
The correct answer is 1. (Actually I got an answer of 8.53 g left, but that's probably just a rounding error).
109.0 g CaCN2 x 1 mole CaCN2/80.11 g CaCN2 x 1 mol CaCO3/ 1 mol CaCN2 = 1.361 mol CaCO3
82.00 g H2O x 1 mol H2O/18.0 g H2O x 1 mol CaCO3/ 3 mol H2) = 1.519 mol CaCO3
Since calcium cyanamide produces fewer moles of CaCO3 than water does, it's the limiting reactant. The 80.5% yield is a red herring, because it has no bearing on solving the problem (so you ignore it).
109.0g CaCN2 x 1 mol CaCN2/80.11 g CaCN2 x 3 mol H2O/1 mol CaCN2 x 18.0 g H2O / 1 mole H2O =
73.474 g H2O 82.000 g H2O - 73.474 g H2O = 8.526 g H2O excess