CaCN2 + 3H2O --> CaCO3 + 2NH3: How much CaCO3 is produced upon reaction of 45g CaCN2 and 45g of H2O? HelpPLZ

Question

Consider the reaction of CaCN2 and water to produce CaCO3 and NH3 according to the reaction

CaCN2 + 3H2O --> CaCO3 + 2NH3 :

How much CaCO3 is produced upon reaction of 45 g CaCN2 and 45 g of H2O?

1. 28 g
2. 19 g
3. 56 g
4. 250 g
5. 83 g
6. 38 g
7. 750 g

Answer 1

The molecular mass of CaCN2: 80
The molecular mass of water: 18
The molecular mass of CaCO3: 100
45g CaCN2 <==> (45/80) Mol CaCN2
45g of H2O <==> (5/2) Mol H2O
One Mol of CaCN2 reacts with 3 Mol of H2O <==> CaCN2 is the limiting chemical.
(45/80) Mol CaCO3 <==> 225/4(g) CaCO3
Answer #3: 56 g

Answer 2

This is a limiting reactant problem. You don't know which reactant is in excess. Convert both 45 g of CaCN2 and 45 g of H2O to moles. Then use the mole ratio from the balanced equation to find out how many moles of CaCO3 is produced from each reactant. Whichever produces the least number of moles of CaCO3 is the limiting reactant, and this is how much CaCO3 is actually produced. Convert this to g of CaCO3 and you have your answer.

Answer 3

since both your reactants have a mass, you must find the limiting reactant.

45 g H2O / 18.015 g H2O = 2.4979 mol H2O
45 g CaCN2 / 80.103 g CaCN2 = .5618 mol CaCN2

2.4979 mol H2O / .5618 mol CaCN2 = 4.44
.5618 mol CaCN2 / .5618 mol CaCN2 = 1.00
in the eqn. you need 3 mol of H2O for every mole of CaCN2 so CaCN2 is your limiting reactant because you have excess H2O.

45 g CaCN2 * 1 mol CaCN2 / 80.103 g CaCN2 * 1 mol CaCO3 / 1 mol CaCN2 * 100.086 g CaCO3 / 1 mol CaCO3 = 56.22 g CaCO3, but considering sig figs it would round to 56 g

The answer is 3. 56 g