Answer:
b. 11.90 Liters
Explanation:
- The balanced equation for the mentioned reaction is:
<em>3O₂ + 4Al → 2Al₂O₃,</em>
It is clear that 3.0 moles of O₂ react with 4.0 moles of Al to produce 2.0 Al₂O₃.
- Firstly, we need to calculate the no. of moles (n) of 36.12 g of Al₂O₃:
<em>n = mass/molar mass</em> = (44.18 g)/(101.96 g/mol) = <em>0.4333 mol.</em>
<u><em>using cross multiplication:</em></u>
3.0 mol of O₂ produces → 2.0 mol of Al₂O₃.
??? mol of O₂ produces → 0.4333 mol of Al₂O₃.
<em>∴ The no. of moles of O₂ needed to produce 36.12 grams of Al₂O₃</em> = (3.0 mol)(0.4333 mol)/(2.0 mol) = <em>0.65 mol.</em>
- Now, we can find the volume of O₂ used during the experiment:
We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 1.3 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 0.65 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 290 K).
<em>∴ V = nRT/P </em>= (0.65 mol)(0.0821 L.atm/mol.K)(290 K)/(1.3 atm) = <em>11.9 L.</em>
<em>So, the right choice is: b. 11.90 Liters.</em>
