Carbon tetrachloride, ccl4, once used as a cleaning fluid and as a fire extinguisher, is produced by heating methane and chlorine. what mass of ch4 is needed to exactly combine with 33.4 g of cl2?

ch4 + 4cl2→ ccl4 + hcl

1.882 g

Explanation:

Data Given

mass of Cl₂ = 33.4 g

mass of CH₄ = ?

Reaction Given:

                       CH₄+ 4Cl₂ → CCl₄ + HCl

Solution:

First find the mass of CH₄ from the reaction that it combine with how many grams of Chlorin.

Look at the balanced reaction

                    CH₄     +     4Cl₂ → CCl₄ + 4HCl

                    1 mol        4 mol

So 1 mole of CH₄ combine with 4 moles of Cl₂

Now

convert the moles into mass for which we have to know molar mass of CH₄ and Cl₂

Molar mass of Cl₂ = 2 (35.5)

Molar mass of  Cl₂  = 71 g/mol

mass of Cl₂

                mass in grams = no. of moles x molar mass

                mass of Cl₂ = 4 mol x 71 g/mol

                mass of Cl₂  = 284 g

Molar mass of CH₄= 12+ 4(1)

Molar mass of CH₄= 16 g/mol

mass of CH₄

                mass in grams = no. of moles x molar mass

                mass of CH₄= 1 mol x 16 g/mol

                mass of CH₄ = 16 g

So,

284 g of Cl₂  combine with 16 g of methane ( CH₄ ) then how many grams of CH₄ is needed to combine with 33.4 g of Cl₂  

Apply unity Formula

                           284 g of Cl₂  ≅ 16 g of methane ( CH₄ )

                           33.4 g of Cl₂  ≅ X g of methane ( CH₄ )

By cross multiplication

                          X g of methane ( CH₄ ) = 16 g x 33.4 g / 284 g

                          X g of methane ( CH₄ ) = 1.88 g

1.882 g of methane (CH₄) will needed to combine with 33.4 g of Cl₂

So

methane (CH₄) = 1.882 g