Question

Consider the following reaction: 2 NO(g) + 5 H2(g) → 2 NH3(g) + 2 H2O(g) Which set of solution maps would be needed to calculate the maximum amount of ammonia (NH3), in grams, that can be synthesized from 45.8 g of nitrogen monoxide (NO) and 12.4 g of hydrogen (H2)? I. g NO → mol NO → mol NH3 → g NH3 II. g H2 → mol H2 → mol NH3 → g NH3 III. g NO → mol NO → mol H2O → g H2O IV. g H2 → mol H2 → mol H2O → g H2O

Answer 1

Answer : The correct option is, (I)
`gNO\rightarrow molNO\rightarrow molNH_3\rightarrow gNH_3`

Solution : Given,

Mass of NO = 45.8 g

Mass of
`H_2` = 12.4 g

Molar mass of NO = 30 g/mole

Molar mass of
`H_2` = 2 g/mole

Molar mass of
`NH_3` = 17 g/mole

First we have to calculate the moles of NO and
`O_2`.

`\text{ Moles of }NO=\frac{\text{ Mass of }NO}{\text{ Molar mass of }NO}=(45.8g)/(30g/mole)=1.53moles`

`\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=(12.4g)/(2g/mole)=6.20moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`2NO(g)+5H_2(g)\rightarrow 2NH_3(g)+2H_2O(g)`

From the balanced reaction we conclude that

As, 2 mole of
`NO` react with 5 mole of
`H_2`

So, 1.53 moles of
`NO` react with
`(1.53)/(2)* 5=3.82` moles of
`H_2`

From this we conclude that,
`H_2` is an excess reagent because the given moles are greater than the required moles and
`NO` 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, 2 mole of
`NO` react to give 2 mole of
`NH_3`

So, 1.53 mole of
`NO` react to give 1.53 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`

`\text{ Mass of }NH_3=(1.53moles)* (17g/mole)=26.0g`

Therefore, the maximum mass of
`NH_3` produced 26.0 grams.