Question

How many liters of oxygen gas, at standard temperature and pressure, will react with 25.0 grams of magnesium metal? Show all of the work used to solve this problem.
2 Mg + O2 ----> 2 MgO

Answer 1

Given:
2 Mg + O2 → 2 MgO 
So,
(25.0 g of Mg / 24.3051 g/mole) x (1/2) x (22.4 L/mole) = 11.5 L  of Oxygen will react with 25 grams of magnesium metal.

Answer 2

Answer : The volume of oxygen gas is, 11.67 liters

Explanation : Given,

Mass of Mg = 25 g

Molar mass of Mg = 24 g/mole

First we have to calculate the moles of Mg.

$\text{Moles of }Mg=\frac{\text{Mass of }Mg}{\text{Molar mass of }Mg}=\frac{25g}{24g/mole}=1.042moles$

Now we have to calculate the moles of oxygen gas.

The balanced chemical reaction is,

$2Mg+O_2\rightarrow 2MgO$

From the balanced chemical reaction, we conclude that

As, 2 mole of Mg react with 1 mole of oxygen gas

So, 1.042 mole of Mg react with $\frac{1.042}{2}=0.521$ mole of oxygen gas

Now we have to calculate the volume of oxygen gas.

As we know that at STP,

1 mole of oxygen gas contains 22.4 L volume of oxygen gas

So, 0.521 mole of oxygen gas contains $0.521\times 22.4=11.67L$ volume of oxygen gas

Therefore, the volume of oxygen gas is, 11.67 liters