The maximum amount of ch3cl that can be prepared by reacting 20 g of ch4 with 10 g cl2 is
Answers
The maximum amount of CH₃Cl that can be prepared by reacting methane and chlorine is 7.11g.
We have to find the maximum amount of CH₃Cl that can be prepared by reacting 20g of CH₄ with 10g Cl₂.
Before solving it, we have to find out which one is limiting reagent.
Chemical reaction of CH₄ and Cl₂ is given by,
CH₄ + Cl₂ ⇒ CH₃Cl + HCl
Here, it is clear that one mole of methane reacts with one mole of chlorine.
- molecular weight of methane = 16g/mol
- molecular weight of chlorine = 71g/mol
∴ 16 g of methane reacts with 71g of chlorine.
∴ 20g of methane reacts with 71/16 × 20 = 88.75g of chlorine. but here only 10g of chlorine. hence, Chlorine is the limiting reagent.
Now chlorine gets the priority of the reaction.
∵ one mole of chlorine produces one mole of CH₃Cl.
∴ 71g of chlorine produces 50.5g of CH₃Cl.
∴ 10g of chlorine produces 50.5/71 × 10 = 7.11g of CH₃Cl.
Therefore the maximum amount of CH₃Cl that can be produced is 7.11g
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