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A sample of 1.778 moles of methane [tex]$\left( \text{CH}_4 \right)$[/tex] is mixed with 1.162 moles of chlorine. What is the theoretical yield, in moles, of [tex]$\text{CCl}_4$[/tex] that can be made from these amounts?

[tex]$
\text{CH}_4(g) + 4 \text{Cl}_2(g) \longrightarrow \text{CCl}_4(l) + 4 \text{HCl}(g)
$[/tex]

Sagot :

To determine the theoretical yield of [tex]\( \text{CCl}_4 \)[/tex] from the given amounts of reactants, we need to analyze the reaction and identify the limiting reactant. The balanced chemical equation is:

[tex]\[ \text{CH}_4(g) + 4 \, \text{Cl}_2(g) \longrightarrow \text{CCl}_4(l) + 4 \, \text{HCl}(g) \][/tex]

Here are the steps to find the theoretical yield:

1. Write the molar ratios based on the balanced equation:
- 1 mole of [tex]\( \text{CH}_4 \)[/tex] reacts with 4 moles of [tex]\( \text{Cl}_2 \)[/tex].
- This produces 1 mole of [tex]\( \text{CCl}_4 \)[/tex].

2. Identify the moles of each reactant:
- We have 1.778 moles of [tex]\( \text{CH}_4 \)[/tex].
- We have 1.162 moles of [tex]\( \text{Cl}_2 \)[/tex].

3. Determine the limiting reactant:
- According to the balanced equation, 1 mole of [tex]\( \text{CH}_4 \)[/tex] requires 4 moles of [tex]\( \text{Cl}_2 \)[/tex].
- The actual requirement of [tex]\( \text{Cl}_2 \)[/tex] for 1.778 moles of [tex]\( \text{CH}_4 \)[/tex] is [tex]\( 1.778 \times 4 \, \text{moles} = 7.112 \, \text{moles} \)[/tex].
- We only have 1.162 moles of [tex]\( \text{Cl}_2 \)[/tex], which is much less than 7.112 moles, indicating that [tex]\( \text{Cl}_2 \)[/tex] is the limiting reactant.

4. Calculate the theoretical yield of [tex]\( \text{CCl}_4 \)[/tex]:
- From the stoichiometry of the balanced equation, 4 moles of [tex]\( \text{Cl}_2 \)[/tex] produces 1 mole of [tex]\( \text{CCl}_4 \)[/tex].
- Therefore, [tex]\( \text{1 mole of} \, \text{Cl}_2 \)[/tex] will produce [tex]\( \frac{1}{4} \, \text{of a mole of} \, \text{CCl}_4 \)[/tex].
- Given 1.162 moles of [tex]\( \text{Cl}_2 \)[/tex], the theoretical yield of [tex]\( \text{CCl}_4 \)[/tex] is:
[tex]\[ \text{Theoretical yield} = \frac{1.162 \, \text{moles}}{4} = 0.2905 \, \text{moles} \][/tex]

Hence, the theoretical yield of [tex]\( \text{CCl}_4 \)[/tex] is 0.2905 moles.