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

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Hussain514 Hussain514 · Answer 1 · 2016-09-15T00:11:03+02:00

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.

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BarrettArcher BarrettArcher · Answer 2 · 2019-06-14T13:13:47+02:00

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.

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

Now we have to calculate the moles of oxygen gas.

The balanced chemical reaction is,

[tex]2Mg+O_2\rightarrow 2MgO[/tex]

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 [tex]\frac{1.042}{2}=0.521[/tex] 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 [tex]0.521\times 22.4=11.67L[/tex] volume of oxygen gas

Therefore, the volume of oxygen gas is, 11.67 liters