Consider the following reaction:
CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g)
What mass of CCl₄ will be formed if 1.4moles of methane react with 1.4 moles of chlorine?

Question

Consider the following reaction: CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g) What mass of CCl₄ will be formed if 1.4moles of methane react with 1.4 moles of chlorine?

Quizplus · Accepted Answer

1. Write the balanced chemical equation:
CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
2. Calculate the limiting reagent:
For 1.4 moles of CH4, we need 4*1.4 = 5.6 moles of Cl2. However, we only have 1.4 moles of Cl2. Therefore, Cl2 is the limiting reagent.
3. Calculate the moles of CCl4 produced:
From the balanced chemical equation, we see that 1 mole of CH4 produces 1 mole of CCl4. Therefore, 1.4 moles of CH4 will produce 1.4 moles of CCl4.
4. Convert moles of CCl4 to grams:
1.4 moles of CCl4 = 1.4 x 154 = 215.6 g = 53.8 g (rounded to one decimal place)

Consider the Following Reaction