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 $O_2$.

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

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

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 $\frac{1.53}{2}\times 5=3.82$ 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 $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

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

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