# Question 66b6c

Jul 15, 2017

$\text{gas added} = 11.9$ $\text{mol}$

#### Explanation:

To solve this, we can use the quantity-volume relationship of gases.

The number of moles of a gas is directly to proportional to the volume the gas occupies.

And, under constant temperature and pressure conditions, we get the equation

$\frac{{V}_{1}}{{n}_{1}} = \frac{{V}_{2}}{{n}_{2}}$ (constant temperature and pressure)

Our known quantities are:

• ${V}_{1}$: $8.15$ $\text{L}$

• ${n}_{1}$: $9.51$ $\text{mol}$

• ${V}_{2}$: $18.3$ $\text{L}$

We can plug these into the equation and solve for the final quantity, ${n}_{2}$:

n_2 = (n_1V_2)/(V_1) = ((9.51color(white)(l))(18.3cancel("L")))/(8.15cancel("L")) = color(red)(21.4 color(red)("mol"

We're asked to find the number of moles needed to be added: so we subtract the initial amount from the final amount:

"gas added" = color(red)(21.4 color(red)("mol" $- 9.51$ $\text{mol}$ = color(blue)(11.9 color(blue)("mol"#