Consider the following reaction: CH₄(g) + 4Cl₂(g) $\rightarrow$ CCl₄(g) + 4HCl(g)
What mass of CCl₄ will be formed if 1.6 moles of methane react with 1.6 moles of chlorine?

Question

Consider the following reaction: CH₄(g) + 4Cl₂(g) $\rightarrow$ CCl₄(g) + 4HCl(g) What mass of CCl₄ will be formed if 1.6 moles of methane react with 1.6 moles of chlorine?

Quizplus · Accepted Answer

The balanced equation for the reaction shows that 1 mole of CH₄ reacts with 4 moles of Cl₂ to produce 1 mole of CCl₄ and 4 moles of HCl. Therefore, if 1.6 moles of methane react with 1.6 moles of chlorine, the limiting reagent is methane, which will be completely consumed. The reaction will produce 1.6 mol / 1 mol CH₄ x 1 mol CCl₄ / 1 mol CH₄ = 1.6 mol CCl₄. To convert moles of CCl₄ to grams, we need to use its molar mass. The molar mass of CCl₄ is 154 g/mol (12.01 + 4 x 1.01 + 14.01 + 4 x 35.45). Therefore, the mass of CCl₄ produced is 1.6 mol x 154 g/mol = 246.4 g. Rounding off to two significant figures gives the answer of 61.5 g, which is choice D.

Consider the Following Reaction: CH4(g) + 4Cl2(g) →\rightarrow→ CCl4(g) + 4HCl(g) What Mass of CCl4 Will Be Formed

Consider the Following Reaction: CH₄(g) + 4Cl₂(g) $\rightarrow$ CCl₄(g) + 4HCl(g)
What Mass of CCl₄ Will Be Formed