Consider the reaction of cacn2 and water to produce caco3 and nh3: cacn2 + 3 h2o → caco3 + 2 nh3 . how much nh3 is produced upon reaction of 65.0 g of cacn2 and 4.0 moles of h2o? 1. 8 mol 2. 0.8 mol 3. 2.7 mol 4. 6.0 mol 5. 1.6 mol

The correct answer is Option 5.

Explanation:

To calculate the number of moles, we use the equation:

Given mass of calcium cyanamide = 65 g

Molar mass of calcium cyanamide = 80.1 g/mol

Putting values in above equation, we get:

Moles of water = 4.0 moles

The chemical equation for the reaction of calcium cyanide and water follows:

By Stoichiometry of the reaction:

1 mole of calcium cyanamide reacts with 3 moles of water.

So, 0.811 moles of calcium cyanamide will react with = of water.

As, given amount of water is more than the required amount. So, it is considered as an excess reagent.

Thus, calcium cyanamide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of calcium cyanamide produces 2 moles of ammonia.

So, 0.811 moles of calcium cyanamide will produce = of ammonia.

Hence, the correct answer is Option 5.