Answer:
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
