Question

What is the theoretical yield of ammonia that can be obtained from the reaction of 10.0 g of H2 and excess N2?

N2 + 3H2 → 2NH3
a. 90.0 g
b. 97.1 g
c. 56.3 g
d. 48.6 g
e. 28.4 g

Answer 1

Theoretical yield is the ideal number, so we'll assume it all reacted without issue.

`(10.0gH _(2) )/(1)`×
`(1molH_(2) )/(2.016gH _(2) )`×
`(2molNH _(3) )/(3molH _(2) )`×
`(17.03gNH _(3) )/(1molNH _(3) )`=56.3gNH↓3

Answer 2

Answer: The correct answer is Option c.

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)

Given mass of hydrogen gas = 10.0 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

`\text{Moles of hydrogen gas}=(10.0g)/(2g/mol)=5.0mol`

The given chemical equation follows:

`N_2+3H_2\rightarrow 2NH_3`

As, nitrogen gas is present in excess. It is considered as an excess reagent.

Hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of hydrogen gas produces 2 moles of ammonia

So, 5.0 moles of hydrogen gas will produce =
`(2)/(3)* 5.0=3.33mol` of ammonia

Now, calculating the mass of ammonia by using equation 1, we get:

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 3.33 moles

Putting values in equation 1, we get:

`3.33mol=\frac{\text{Mass of ammonia}}{17g/mol}\\\\\text{Mass of ammonia}=(3.33mol* 17g/mol)=56.3g`

Hence, the correct answer is Option c.