Question d0d53

Apr 16, 2015

You'd need 30 L of oxygen gas under those conditions of temperature and pressure to react with that much aluminium.

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

Notice that you have a $\textcolor{red}{4} : 3$ mole ratio between aluminium and oxygen, which means that, regardless of how many moles of the former that react, you'll always have 3/4 less moles of oxygen required for the reaction to take place.

So, you know the mass of aluminium that reacts, which means that you can determine the number of moles that react by using aluminium's molar mass

51.7cancel("g") * "1 mole Al"/(26.982cancel("g")) = "1.916 moles Al"

Use the mole ratio to determine how many moles of oxygen you need

1.916cancel("moles Al") * ("3 moles "O_2)/(color(red)(4)cancel("moles Al")) = "1.437 moles "# ${O}_{2}$

All you have to do now is use the ideal gas law equation to solve for the volume you need

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

$V = \left(1.437 \cancel{\text{moles") * 0.082(cancel("atm") * "L")/(cancel("mol") * cancel("K")) * (273.15 + 20)cancel("K"))/(777/760cancel("atm}}\right)$

${V}_{{O}_{2}} = \text{33.78 L}$

SIDE NOTE Do not forget to convert degrees Celsius and mmHg to Kelvin and atm, respectively.

Rounded to one sig fig, the number of sig figs given for the temperature, the answer will be

${V}_{{O}_{2}} = \textcolor{g r e e n}{\text{30 L}}$