# Question #48338

Oct 12, 2015

1.52 mols of NO2.

#### Explanation:

So EqK=3.2 and let x be the NO2 amount.

So 3.2= (1.2)(1.2) / (2.6-1.2)(x)
And now solve for x..

You will arrive at x=0.3214...
Add 1.2 as it the amount of the product needed, so..

1.2+0.3214...=1.52 mols. Correct to 3 significant figures.

Oct 12, 2015

We will need $1.52 m o l$ of $N {O}_{2}$ to react with $2.6 m o l$ of $S {O}_{2}$ to produce $1.2 m o l$ of $S {O}_{3}$.

#### Explanation:

We will solve this question through ICE table.

Let us write the equilibrium and list the initial, change and concentrations at equilibrium.

We will assume 1 litre volume so we can calculate the concentrations.

$\text{ " " " " " " " " SO_2(g)+ NO_2(g)" " "rightleftharpoons" } S {O}_{3} \left(g\right) + N O \left(g\right)$
$I n i t i a l \text{ " " " "2.6M " " " " " "xM" " " " " " " " " " "0M" " " } 0 M$
$\text{Change" " " " "-1.2M" " -1.2M " " " " " "+1.2M" " } + 1.2 M$
$E q u i l i b r i u m \text{ "1.4M" " " (x-1.2)M" " " " " " "1.2M" " " } 1.2 M$

The expression of $K$ would be:

${K}_{C} = \frac{\left[S {O}_{3}\right] \left[N O\right]}{\left[S {O}_{2}\right] \left[N {O}_{2}\right]} = 3.2$

Replace the equilibrium concentrations by their values in the expression of ${K}_{C}$ we get:

${K}_{C} = \frac{\left(1.2 M\right) \left(1.2 M\right)}{\left(1.4 M\right) \left(x - 1.2 M\right)} = 3.2$

Solving for $x$ we get: $x = 1.53 M$.

Therefore, we will need $1.52 m o l$ of $N {O}_{2}$ to react with $2.6 m o l$ of $S {O}_{2}$ to produce $1.2 m o l$ of $S {O}_{3}$.