Questions: A sample of 5.00 g of methane (CH4) is mixed with 14.5 g of chlorine (Cl2). a. Determine which is the limiting reactant according to the following equation: CH4(g) + 4 Cl2(g) → CCl4(l) + 4 HCl(g) b. The limiting reactant is . - What is the maximum mass of CCl4 that can be formed? Maximum mass = g

A sample of 5.00 g of methane (CH4) is mixed with 14.5 g of chlorine (Cl2).
a. Determine which is the limiting reactant according to the following equation:
CH4(g) + 4 Cl2(g) → CCl4(l) + 4 HCl(g)
b. The limiting reactant is .
- What is the maximum mass of CCl4 that can be formed?

Maximum mass = g

Transcript text: A sample of 5.00 g of methane $\left(\mathrm{CH}_{4}\right)$ is mixed with 14.5 g of chlorine $\left(\mathrm{Cl}_{2}\right)$. a. Determine which is the limiting reactant according to the following equation: \[ \mathrm{CH}_{4}(\mathrm{~g})+4 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CCl}_{4}(\ell)+4 \mathrm{HCl}(\mathrm{g}) \] b. The limiting reactant is $\square$ . - What is the maximum mass of $\mathrm{CCl}_{4}$ that can be formed? Maximum mass $=$ $\square$ g

Solution

Solution Steps

Step 1: Calculate Moles of Each Reactant

First, we need to calculate the number of moles of each reactant.

For methane ($\mathrm{CH}_4$): \[ \text{Molar mass of } \mathrm{CH}_4 = 12.01 + 4 \times 1.008 = 16.042 \, \text{g/mol} \] \[ \text{Moles of } \mathrm{CH}_4 = \frac{5.00 \, \text{g}}{16.042 \, \text{g/mol}} = 0.3116 \, \text{mol} \]

For chlorine ($\mathrm{Cl}_2$): \[ \text{Molar mass of } \mathrm{Cl}_2 = 2 \times 35.453 = 70.906 \, \text{g/mol} \] \[ \text{Moles of } \mathrm{Cl}_2 = \frac{14.5 \, \text{g}}{70.906 \, \text{g/mol}} = 0.2045 \, \text{mol} \]

Step 2: Determine the Limiting Reactant

The balanced chemical equation is: \[ \mathrm{CH}_4 + 4 \mathrm{Cl}_2 \rightarrow \mathrm{CCl}_4 + 4 \mathrm{HCl} \]

From the equation, 1 mole of $\mathrm{CH}_4$ reacts with 4 moles of $\mathrm{Cl}_2$.

Calculate the required moles of $\mathrm{Cl}_2$ for the available $\mathrm{CH}_4$: \[ \text{Required moles of } \mathrm{Cl}_2 = 0.3116 \, \text{mol} \times 4 = 1.2464 \, \text{mol} \]

Since we only have 0.2045 moles of $\mathrm{Cl}_2$, $\mathrm{Cl}_2$ is the limiting reactant.

$\boxed{\text{The limiting reactant is } \mathrm{Cl}_2}$

Step 3: Calculate the Maximum Mass of $\mathrm{CCl}_4$

Using the limiting reactant ($\mathrm{Cl}_2$), we calculate the maximum mass of $\mathrm{CCl}_4$ that can be formed.

From the balanced equation, 4 moles of $\mathrm{Cl}_2$ produce 1 mole of $\mathrm{CCl}_4$.

Calculate the moles of $\mathrm{CCl}_4$ that can be formed: \[ \text{Moles of } \mathrm{CCl}_4 = \frac{0.2045 \, \text{mol} \, \mathrm{Cl}_2}{4} = 0.05113 \, \text{mol} \]

Calculate the mass of $\mathrm{CCl}_4$: \[ \text{Molar mass of } \mathrm{CCl}_4 = 12.01 + 4 \times 35.453 = 153.823 \, \text{g/mol} \] \[ \text{Mass of } \mathrm{CCl}_4 = 0.05113 \, \text{mol} \times 153.823 \, \text{g/mol} = 7.864 \, \text{g} \]

$\boxed{\text{Maximum mass} = 7.864 \, \text{g}}$