Question

Ammonia gas is formed from nitrogen gas and hydrogen gas, according to the following equation, N2 (g) + 3H2 (g) 2NH3 (g). If 140 grams of nitrogen gas is allowed to react with an excess of hydrogen gas to produce 155 grams of ammonia, what is the percent yield of this reaction?

Answer 1

Answer: The percent yield of the reaction is 91.20 %.

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 nitrogen gas = 140 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

`\text{Moles of nitrogen gas}=(140g)/(28g/mol)=5mol`

For the given chemical reaction:

`N_2(g)+3H_2(g)\rightarrow 2NH_3(g)`

As, hydrogen is present in excess. Thus, it is considered as an excess reagent. Nitrogen gas is considered as limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of nitrogen gas produces 2 moles of ammonia.

So, 5 moles of nitrogen gas will produce =
`(2)/(1)* 5=10mol` of ammonia.

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

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 10 moles

Putting values in above equation, we get:

`10mol=\frac{\text{Mass of ammonia}}{17g/mol}\\\\\text{Mass of ammonia}=170g`

To calculate the percentage yield of the reaction, we use the equation:

`\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

Experimental yield of ammonia = 155 g

Theoretical yield of ammonia = 170 g

Putting values in above equation, we get:

`\%\text{ yield of ammonia}=(155g)/(170g)* 100\\\\\% \text{yield of ammonia}=91.20\%`

Hence, the percent yield of the reaction is 91.20 %.