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>
Answer:
C3H6O
Explanation:
The percentage composition of the elements in the compound are given as follows:
62.1 % carbon = 62.1g of C
10.5 % hydrogen = 10.5g of H
27.6 % oxygen = 27.6g of O
Next, we convert each mass to mole by dividing by their molar/atomic mass
C = 62.1/12 = 5.175mol
H = 10.5/1 = 10.5mol
O = 27.6/16 = 1.725mol
Next, we divide each mole value by the smallest mole value (1.725)
C = 5.175mol ÷ 1.725 = 3
H = 10.5mol ÷ 1.725 = 6.086
O = 1.725mol ÷ 1.725 = 1
The empirical ratio approximately of C:H:O is 3:6:1, hence, the empirical formula is C3H6O