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How many liters of oxygen gas can react with 84.0 grams of lithium metal at standard temperature and pressure? Show all of the work used to find your answer. 4Li + O2 yields 2Li2O

Respuesta :

From the equation:
4mol Li react with 1 mol O2
Molar mass Li = 7g/mol
mol in 84g Li = 84/7 = 12 mol Li
From the equation - 12 mol Li will react with 3 mol O2

At STP 1 mol O2 has volume = 22.4L
At STP 3 mol O2 has volume = 3*22.4 = 67.2L O2 gas will react.

Answer: The volume of oxygen gas reacted is 67.2 L.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

For Lithium

Given mass of lithium = 84 g

Molar mass of lithium = 7 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of lithium}=\frac{84g}{7g/mol}=12mol[/tex]

For the given chemical reaction:

[tex]4Li+O_2\rightarrow 2Li_2O[/tex]

By Stoichiometry of the reaction:

4 moles of lithium reacts with 1 mole of oxygen gas.

So, 12 moles of lithium metal will react with = [tex]\frac{1}{4}\times 12=3moles[/tex] of oxygen gas.

At STP:

1 mole of a gas occupies 22.4 L of gas.

So, 3 moles of a gas will occupy = [tex]3\times 22.4=67.2L[/tex]

Hence, the volume of oxygen gas reacted is 67.2 L.

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