Respuesta :
ANSWER
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
[tex]\text{ 4Al}_{(s)}\text{ + 3O}_{2(g)}\text{ }\rightarrow\text{ 2Al}_2O_3[/tex]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
[tex]\text{ mole }=\text{ }\frac{\text{ mass}}{\text{ molar mass}}[/tex]Recall, that the molar mass of Al is 26.98 g/mol
[tex]\begin{gathered} \text{ mole }=\text{ }\frac{37.86}{26.98} \\ \text{ mole }=\text{ 1.403 moles} \end{gathered}[/tex]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
[tex]\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{ }\times\text{ 4 mol Al }=\text{ 3 mol of O}_2\text{ }\times\text{ 1.403 mol Al} \\ \text{ x mol 4mol Al }=\frac{3\text{ mol of O}_2\text{ }\times1.403\cancel{mol\text{ Al}}}{4\cancel{mol\text{ Al}}} \\ \text{ x 4 mol Al }=\text{ }\frac{3\text{ }\times\text{ 1.403}}{4} \\ \text{ x }=\frac{4.209}{4} \\ \text{ x = 1.0523 mol} \end{gathered}[/tex]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
[tex]\begin{gathered} \text{ 1 mole }\rightarrow\text{ 22.4 mol/L} \\ \text{ 1.0523 mole }\rightarrow\text{ xL} \\ \text{ cross multiply} \\ \text{ 1 }\times\text{ x }=\text{ 22.4 }\times\text{ 1.0523} \\ \text{ x }=\text{ 23.57152 L} \end{gathered}[/tex]Hence, the volume of oxygen gas at STP IS 23. 57152L
