Question

1.46 g H2 is allowed to react with 10.5 g N2, producing 2.72 g NH3. What is the theoretical yield in grams for this reaction under the given conditions?

Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer : The theoretical yield in grams for this reaction under the given conditions is, 8.28 grams.

Explanation : Given,

Mass of
`H_2` = 1.46 g

Mass of
`N_2` = 10.5 g

Molar mass of
`H_2` = 2 g/mol

Molar mass of
`N_2` = 28 g/mol

First we have to calculate the moles of
`H_2` and
`N_2`.

`\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}`

`\text{Moles of }H_2=(1.46g)/(2g/mol)=0.73mol`

and,

`\text{Moles of }N_2=\frac{\text{Given mass }N_2}{\text{Molar mass }N_2}`

`\text{Moles of }N_2=(10.5g)/(28g/mol)=0.375mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

`3H_2+N_2\rightarrow 2NH_3`

From the balanced reaction we conclude that

As, 3 mole of
`H_2` react with 1 mole of
`N_2`

So, 0.73 moles of
`H_2` react with
`(0.73)/(3)=0.243` moles of
`N_2`

From this we conclude that,
`N_2` is an excess reagent because the given moles are greater than the required moles and
`H_2` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`NH_3`

From the reaction, we conclude that

As, 3 mole of
`H_2` react to give 2 mole of
`NH_3`

So, 0.73 mole of
`H_2` react to give
`(2)/(3)* 0.73=0.487` mole of
`NH_3`

Now we have to calculate the mass of
`NH_3`

`\text{ Mass of }NH_3=\text{ Moles of }NH_3* \text{ Molar mass of }NH_3`

Molar mass of
`NH_3` = 17 g/mole

`\text{ Mass of }NH_3=(0.487moles)* (17g/mole)=8.28g`

Therefore, the theoretical yield in grams for this reaction under the given conditions is, 8.28 grams.