Question

Ammonia reacts with ${\mathrm{O}}_2$to form either $\mathrm{NO}\left(\mathrm{g}\right)$or ${\mathrm{NO}}_2\left(\mathrm{g}\right)$according to these unbalanced equations:

${\mathrm{NH}}_3\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to \mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

${\mathrm{NH}}_3\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{NO}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

In a certain experiment, 2.00 moles of ${\mathrm{NH}}_3\left(\mathrm{g}\right)$and 10.00 moles of ${\mathrm{O}}_2\left(\mathrm{g}\right)$ are contained in a closed flask. After the reaction is complete, 6.75 moles of ${\mathrm{O}}_2\left(\mathrm{g}\right)$remains. Calculate the number of moles of ${\mathrm{NO}}_2\left(\mathrm{g}\right)$ in the product mixture. (Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.)

Short answer

Thenumber of moles of${\mathrm{NO}}_2\left(\mathrm{g}\right)$ in the product mixture is$0.50\mathrm{moles}$.

Step-by-step solution

Definition

A balanced chemical equation is one in which the number of atoms present in the reactant side of the equation equals the number of atoms present in the product side of the equation.

Balanced chemical reactions

Here the balanced chemical reactions are,

$4{\mathrm{NH}}_3\left(\mathrm{g}\right)+5{\mathrm{O}}_2\left(\mathrm{g}\right)\to 4\mathrm{NO}\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)-----\left(1\right)$

$4{\mathrm{NH}}_3\left(\mathrm{g}\right)+7{\mathrm{O}}_2\left(\mathrm{g}\right)\to 4{\mathrm{NO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)-----\left(2\right)$

So, Number of moles of oxygen reactedrole="math" localid="1650343593307" $=10.00-6.75$

role="math" localid="1650343616546" $=3.25$Moles

If x is the number of moles of oxygen used in reaction (1) and y is the number of moles of oxygen consumed in reaction (2),

$\mathrm{x}+\mathrm{y}=3.25-----\left(3\right)$

For5 moles${\mathrm{O}}_2$4 moles ammoniais consumed from equation (1),

Lets determine x moles of ${\mathrm{O}}_2$$\left(\frac{\mathrm{x}}{5}\right)4=\left(\frac{4}{5}\right)\mathrm{x}$moles Ammonia is consumed.

For 7 moles ${\mathrm{O}}_2$4 moles ammoniais consumed from equation (2),

Lets determine y moles of ${\mathrm{O}}_2$$\left(\frac{\mathrm{y}}{7}\right)\mathrm{y}=\left(\frac{4}{7}\right)\mathrm{y}$ moles Ammonia is consumed.

Determining the number of moles of NO in the product mixture.

Let’s find the total number of moles of Ammonia consumed

$\left(\frac{4}{5}\right)\mathrm{x}+\left(\frac{\mathrm{y}}{7}\right)4=2.00$

$0.8\mathrm{x}+0.571\mathrm{y}=2.00-----\left(4\right)$

Let’s add $0.571$in equation (4),

$\left[\left(0.571\right)-\left(0.8\right)\right]\mathrm{x}=\left(3.25\right)\left(0.571\right)-2.00$

$\left(-0.229\right)\mathrm{x}=\left(1.855\right)-2.00$

$\mathrm{x}=0.633\mathrm{mol}$

For 5 moles ${\mathrm{O}}_2$4 moles ammonia is consumed from equation (1),

Thus for $0.633\mathrm{mol}\mathrm{of}{\mathrm{O}}_2=\left(\frac{0.633}{5}\right)4$

Therefore,$0.50\mathrm{moles}\mathrm{of}\text{N}{\mathrm{O}}_2$is produced.