Question

Question #48338

Answer 1

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.

Answer 2

We will need `1.52 mol` of `NO_2` to react with `2.6mol` of `SO_2` to produce `1.2mol` of `SO_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.

`" " " " " " " " " SO_2(g)+ NO_2(g)" " "rightleftharpoons" " SO_3(g) + NO(g)`
`Initial" " " " "2.6M " " " " " "xM" " " " " " " " " " "0M" " " "0M`
`"Change" " " " "-1.2M" " -1.2M " " " " " "+1.2M" " "+1.2M`
`Equilibrium" "1.4M" " " (x-1.2)M" " " " " " "1.2M" " " "1.2M`

The expression of `K` would be:

`K_C= ([SO_3][NO])/([SO_2][NO_2])=3.2`

Replace the equilibrium concentrations by their values in the expression of `K_C` we get:

`K_C= ((1.2M)(1.2M))/((1.4M)(x-1.2M))=3.2`

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

Therefore, we will need `1.52 mol` of `NO_2` to react with `2.6mol` of `SO_2` to produce `1.2mol` of `SO_3`.