Question

11. In an experiment 3.25 g of NH3 is reacted with 3.50 g O2 according to the following reaction: NH3 + O2 NO + H2O a. What is the limiting reactant? b. How much (g) excess reactant is left after the reaction is complete? c. What is the theoretical yield (g) of NO? d. What is the percent yield if only 0.98 g of NO were produced

Answer 1

Answer :

(a) The limiting reactant is, $O_2$

(b) The mass of excess reactant is, 1.7646 g

(c) The theoretical yield of NO is, 2.616 g

(d) The percent yield of the reaction is, 37.46 %

Explanation : Given,

Mass of $NH_3$ = 3.25 g

Mass of $O_2$ = 3.50 g

Molar mass of $NH_3$ = 17 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $NO$ = 30 g/mole

First we have to calculate the moles of $NH_3$ and $O_2$.

$\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}=\frac{3.25g}{17g/mole}=0.191moles$

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{3.50g}{32g/mole}=0.109moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

From the balanced reaction we conclude that

As, 5 moles of $O_2$ react with 4 mole of $NH_3$

So, 0.109 moles of $O_2$ react with $\frac{4}{5}\times 0.109=0.0872$ moles of $NH_3$

From this we conclude that, $NH_3$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

The limiting reactant is, $O_2$

Excess moles of $NH_3$ = 0.191 - 0.0872 = 0.1038 mole

Now we have to calculate the mass of $NH_3$.

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

$\text{Mass of }NH_3=(0.1038mole)\times (17g/mole)=1.7646g$

The mass of excess reactant = 1.7646 g

Now we have to calculate the moles of $NO$.

As, 5 moles of $O_2$ react with 4 mole of $NO$

So, 0.109 moles of $O_2$ react with $\frac{4}{5}\times 0.109=0.0872$ moles of $NO$

Now we have to calculate the mass of $NO$.

$\text{Mass of }NO=\text{Moles of }NO\times \text{Molar mass of }NO$

$\text{Mass of }NO=(0.0872mole)\times (30g/mole)=2.616g$

Now we have to calculate the percent yield of $NO$

$\%\text{ yield of }NO=\frac{\text{Actual yield of }NO}{\text{Theoretical yield of }NO}\times 100=\frac{0.98g}{2.616g}\times 100=37.46\%$

The percent yield of the reaction is, 37.46 %