Question

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Answer 1

The reaction is

2CH₄(g) + O₂(g) + 4Cl₂(g) ---> 2CO(g) + 8HCl(g)

So as per the stoichiometry of given equation , 2 moles of methane will react with one mole of O2 and four moles of chlorine molecule to give 8 moles of HCl

Given : oxygen is in excess [so it will not limit the yield of reaction]

Let us calculate moles of methane and Cl2 present

Moles = Mass / molecular weight

(i) 46.0 g CH4, mol wt of CH4 = 12 + 4 = 16 g / mole

Moles of methane = 46 / 16 = 2.875 moles

for these moles of methane , moles of Cl2 required = 2X 2.875 = 5.75 moles

(ii) 131 g Cl2 , molwt of Cl2 = 2 X 35. 5 = 71 g / mole

Moles of Cl2 = 131 / 71 = 1.845 moles

so the limiting reagent is Cl2

now again

4 moles of Cl2 will react with 2 moles of CH4 to give 8 moles of HCl

1 mole of Cl2 will react with 0.5 moles of CH4 to give 2 moles of HCl

1.845 moles of Cl2 will react with 0.9225 moles of CH4 to give 2 X 1.845 moles of Hcl

Moles of HCl produced = 3.69 moles

Mass of HCl produced = Moles X molecular weight = 3.69 X 36.5 = 134.685 grams of HCl