Question

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

Answer 1

Theoretical yield is the ideal number, so we'll assume it all reacted without issue.
$\frac{10.0gH _{2} }{1}$×$\frac{1molH_{2} }{2.016gH _{2} }$×$\frac{2molNH _{3} }{3molH _{2} }$×$\frac{17.03gNH _{3} }{1molNH _{3} }$=56.3gNH↓3

Answer 2

Answer: The correct answer is Option c.

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:

$\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5.0mol$

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 = $\frac{2}{3}\times 5.0=3.33mol$ 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:

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

Hence, the correct answer is Option c.