2. if 20.0 moles of oxygen are burned in the combustion of butane, how many moles of h₂o are produced? (1…

Explanation:

Step1: Balance the butane - combustion equation

The balanced equation for the combustion of butane is (2C_{4}H_{10}+13O_{2}\rightarrow8CO_{2}+10H_{2}O).

Step2: Set up mole - ratio calculation

From the balanced equation, the mole - ratio of (O_{2}) to (H_{2}O) is (13:10). Let (x) be the moles of (H_{2}O) produced. We have the proportion (\frac{13}{10}=\frac{20.0}{x}).

Step3: Solve for (x)

Cross - multiply to get (13x = 10\times20.0), then (x=\frac{10\times20.0}{13}\approx15.4) moles.

Step4: Balance the aluminum chloride decomposition equation

The balanced equation for the decomposition of aluminum chloride is (2AlCl_{3}\rightarrow2Al + 3Cl_{2}).

Step5: Calculate moles of (AlCl_{3})

The molar mass of (AlCl_{3}) is (M = 27.0+3\times35.5=133.5\ g/mol). The moles of (AlCl_{3}), (n=\frac{m}{M}=\frac{120.0\ g}{133.5\ g/mol}\approx0.899) moles.

Step6: Set up mole - ratio calculation for (Cl_{2}) production

From the balanced equation, the mole - ratio of (AlCl_{3}) to (Cl_{2}) is (2:3). Let (y) be the moles of (Cl_{2}) produced. We have the proportion (\frac{2}{3}=\frac{0.899}{y}), so (y=\frac{3\times0.899}{2}=1.3485) moles.

Step7: Calculate mass of (Cl_{2})

The molar mass of (Cl_{2}) is (M_{Cl_{2}}=2\times35.5 = 71\ g/mol). The mass of (Cl_{2}), (m = n\times M=1.3485\ mol\times71\ g/mol\approx95.7\ g)

Answer:

1. Moles of (H_{2}O) produced: approximately (15.4) moles
2. Mass of (Cl_{2}) produced: approximately (95.7\ g)