Question

How many liters of oxygen gas at STP are required to react with 37.86g of aluminum in the production of aluminum oxide?

Answer 1

The volume of oxygen gas at STP IS 23. 57152L

EXPLANATION;

Given that;

The mass of aluminum is 37.86g

To find the volume of oxygen at STP, follow the steps below

Step 1; Write a balanced equation for the reaction

`\text{ 4Al}_((s))\text{ + 3O}_(2(g))\text{ }\rightarrow\text{ 2Al}_2O_3`

In the reaction above, 4 moles of Al reacts with 3 moles of O2

Step 2; Find the number of moles of Al using the formula below

`\text{ mole }=\text{ }\frac{\text{ mass}}{\text{ molar mass}}`

Recall, that the molar mass of Al is 26.98 g/mol

`\begin{gathered} \text{ mole }=\text{ }(37.86)/(26.98) \\ \text{ mole }=\text{ 1.403 moles} \end{gathered}`

Therefore, the number of moles of Al is 1.403 moles

Step 3; Find the moles of O2 using a stoichiometry ratio

4 moles of Al is equivalent to 3 moles of oxygen

Let x represent the number of moles of oxygen

`\begin{gathered} \text{ 4 mol Al }\rightarrow\text{ 3 mole of O}_2 \\ \text{ 1.403 mol Al }\rightarrow\text{ x mole of O}_2 \\ \text{ cross multiply} \\ \text{ x mol O}_2\text{ }*\text{ 4 mol Al }=\text{ 3 mol of O}_2\text{ }*\text{ 1.403 mol Al} \\ \text{ x mol 4mol Al }=\frac{3\text{ mol of O}_2\text{ }*1.403\cancel{mol\text{ Al}}}{4\cancel{mol\text{ Al}}} \\ \text{ x 4 mol Al }=\text{ }\frac{3\text{ }*\text{ 1.403}}{4} \\ \text{ x }=(4.209)/(4) \\ \text{ x = 1.0523 mol} \end{gathered}`

Therefore, the number of moles of oxygen is 1.0523 moles

Step 4; Find the volume of oxygen gas

1 mol of gas at S.T.P is 22.4

`\begin{gathered} \text{ 1 mole }\rightarrow\text{ 22.4 mol/L} \\ \text{ 1.0523 mole }\rightarrow\text{ xL} \\ \text{ cross multiply} \\ \text{ 1 }*\text{ x }=\text{ 22.4 }*\text{ 1.0523} \\ \text{ x }=\text{ 23.57152 L} \end{gathered}`

Hence, the volume of oxygen gas at STP IS 23. 57152L