Respuesta :
The mass of CCl₄ the would be formed if 19.2 g of methane react with 107 g of chlorine gas is 58.0 g
To determine the mass CCl₄ that will be formed,
First, we will write the balanced equation of reaction properly
CH₄(g) +4Cl₂(g) → CCl₄(g) + 4HCl(g)
This means that, 1 mole of CH₄ would react with 4 moles of Cl₂ to yield 1 mole of CCl₄ and 4 moles of HCl
Now, we will determine the number of moles of methane and that of chlorine gas present.
From the question,
mass of methane = 19.2 g
From the formula
[tex]Number \ of \ moles = \frac{Mass}{Molar \ mass}[/tex]
Molar mass of methane = 16.04 g/mol
∴ [tex]Number \ of \ moles \ of \ methane = \frac{19.2}{16.04}[/tex]
[tex]Number \ of \ moles \ of \ methane = 1.197 \ moles[/tex]
For Chlorine
Mass of chlorine = 107 g
Molar mass of chlorine gas = 70.90 g/mol
∴ [tex]Number \ of \ moles \ of \ chlorine = \frac{107}{70.90}[/tex]
[tex]Number \ of \ moles \ of \ chlorine = 1.509 \ moles[/tex]
Now, since 1 mole of methane (CH₄) is required to react with 4 moles of chlorine (Cl₂),
Then,
[tex]\frac{1.509}{4}[/tex] moles of methane will react completely with 1.509 moles of chlorine
∴ Only 0.37725 moles of the methane will react with the 1.509 moles of chlorine.
This means methane is the excess reagent and chlorine is the limiting reagent.
Now, from the balanced chemical equation of the reaction,
1 mole of CH₄ would react with 4 moles of Cl₂ to yield 1 mole of CCl₄
∴ 0.37725 moles of methane will react with the 1.509 moles of chlorine to yield 0.37725 moles of CCl₄
This means 0.37725 moles of CCl₄ was produced during the reaction.
Now, we will determine the mass of CCl₄ produced
From the formula
Mass = Number of moles × Molar mass
Molar mass of CCl₄ = 153.81 g/mol
∴ Mass of CCl₄ produced = 0.37725 × 153.81
Mass of CCl₄ produced = 58.0248
Mass of CCl₄ produced ≅ 58.0 g
Hence, the mass of CCl₄ the would be formed if 19.2 g of methane react with 107 g of chlorine gas is 58.0 g
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