Question

A yield of NH₃ of approximately 98% can be obtained at 200°C and 1,000 atmospheres of pressure.How many grams of N₂ must react to form 1.7 grams of ammonia, NH₃?

a. 0.0058 g
b. .052 g
c. 1.4 g
d. 2.8 g

Answer 1

Answer: The mass of nitrogen gas that must be reacted is 1.4 grams

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}}` .....(1)

• For ammonia:

Given mass of ammonia = 1.7 g

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

`\text{Moles of ammonia}=(1.7g)/(17g/mol)=0.1mol`

The chemical equation for the production of ammonia follows:

`N_2+3H_2\rightarrow 2NH_3`

By Stoichiometry of the reaction:

2 moles of ammonia is produced by 1 mole of nitrogen gas

So, 0.1 moles of ammonia will be produced from
`(1)/(2)* 0.1=0.05mol` of nitrogen gas

Now, calculating the mass of nitrogen gas from equation 1, we get:

Molar mass of nitrogen gas = 28 g/mol

Moles of nitrogen gas = 0.05 moles

Putting values in equation 1, we get:

`0.05mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.05mol* 28g/mol)=1.4g`

Hence, the mass of nitrogen gas that must be reacted is 1.4 grams