# Question 75041

Jun 21, 2015

You'd need 43 L of oxygen to react with that much aluminium.

#### Explanation:

Start by taking a look at the balanced chemical equation

$\textcolor{red}{4} A {l}_{\left(s\right)} + 3 {O}_{2 \left(g\right)} \to 2 A {l}_{2} {O}_{3 \left(s\right)}$

Notice that you have a $\textcolor{red}{4} : 3$ mole ratio between aluminium and oxygen. This means that, in order to get a complete reaction, you need 3/4 less moles of oxygen than you do moles of aluminium.

Since you know how many moles of aluminium react, you get

2.4cancel("moles"Al) * ("3 moles"O_2)/(color(red)(4)cancel("moles"Al)) = "1.8 moles" ${O}_{2}$

You can use the pressure and temperature at which the oxygen is measured to determine what volume that many moles will occupy. To do that, use the ideal gas law equation

$P V = n R T \implies V = \frac{n R T}{P}$

Do not forget to convert the temperature from degrees Celsius to Kelvin and the pressure from mmHg to atm!

V_(O_2) = (1.8cancel("moles") * 0.082(cancel("atm") * "L")/(cancel("mol") * cancel("K")) * (273.15 + 25)cancel("K"))/(782/760cancel("atm")) = "42.77 L"#

Rounded to two sig figs, the number of sig figs you gave for the moles of aluminium, the answer will be

${V}_{{O}_{2}} = \textcolor{g r e e n}{\text{43 L}}$