Question

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

Answer 1

1.6 mol

Step-by-step explanation:

the answer above is correct. I dont know why someone rated it as 1/5 stars bc its all right

Answer 2

Answer: The correct answer is Option 5.

Step-by-step explanation:

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

• For calcium cyanamide:

Given mass of calcium cyanamide = 65 g

Molar mass of calcium cyanamide = 80.1 g/mol

Putting values in above equation, we get:

`\text{Moles of calcium cyanamide}=(65g)/(80.1g/mol)=0.811mol`

Moles of water = 4.0 moles

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

`CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3`

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 =
`(3)/(1)* 0.811=2.433mol` 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 =
`(2)/(1)* 0.811=1.6mol` of ammonia.

Hence, the correct answer is Option 5.