Consider the following reaction: CH₄(g) + 4Cl₂(g) $\rightarrow$ CCl₄(g) + 4HCl(g)
What mass of CCl₄ is formed by the reaction of 8.00 g of methane with an excess of chlorine?

Question

Consider the following reaction: CH₄(g) + 4Cl₂(g) $\rightarrow$CCl₄(g) + 4HCl(g) What mass of CCl₄ is formed by the reaction of 8.00 g of methane with an excess of chlorine?

Quizplus · Accepted Answer

The molar mass of CH₄ is 16.04 g/mol, so 8.00 g of CH₄ is 0.499 mol. The reaction shows a 1:1 molar ratio between CH₄ and CCl₄, so 0.499 mol of CH₄ produces 0.499 mol of CCl₄. The molar mass of CCl₄ is 153.82 g/mol, so 0.499 mol of CCl₄ is 76.7 g.

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

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