Consider the reaction of CaCN2 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? 1. 750 g 2. 250 g 3. 56 g 4. 19 g 5. 83 g 6. 28 g 7. 38 g

56 g. Option 3.

Explanation:

The reaction is: CaCN₂ + 3H₂O → CaCO₃ + 2 NH₃

1 mol of calcium cianide reacts with 3 moles of water in order to produce 1 mol of calcium carbonate and 2 moles of ammonia

We have the mass of each reactant, so let's convert the mass to moles:

45 g. 1mol / 80.08 g = 0.562 moles of cianide

45 g. 1mol / 18 g = 2.5 moles of water

The cianide is the limiting reactant:

3 moles of water need 1 mol of cianide to react

Then, 2.5 moles of water will need (2.5. 1) / 3 = 0.833 moles

As we have 0.562 moles of CN⁻ we don't have enough

We can work now, on the reaction:

Ratio is 1:1. Therefore 0.562 moles of cianide will produce 0.562 moles of carbonate

Let's convert the mass to moles to find the answer:

0.562 mol. 100.08 g / 1 mol = 56.2 g