Question

y. If you have 24.5 L of methane gas reacting with 32.2 L of chlorine gas, predict the product and calculate the amount of product in liters. (2 pts)

Answer 1

This type of question is treated as an ideal condition.

We know that,

Molar volume of gas at STP = 22.4L/mol

Volume of methane = 24.5L

Volume of chlorine = 32.2L

(Number of moles of gass = (volume of gas)/(Molar volume of gas)

(Volume of gas = (number of moles of gas)(molar volume of gas)

step (1) Calculation for number of moles of reactant

Number of moles of methane = (24.5L)/(22.4L/mol)

Number of moles of methane = 1.094mol

Number of moles of chlorine = (32.2L)/(22.4L/mol)

Number of moles chlorine = 1.437mol

Step(2) calculation for number of moles of products

Chemical reaction is

CH_{​​​​​4}​(g) + Cl_{​​​​​2} (g) ----------------> CH₃Cl(g) + HCl(g)

From the stoichiometry of chemical reaction

1 mol methane gas react with 1 mol chlorine and produced 1mol chloromethane(CH₃Cl) and 1mol HCl gas

So,

1.094 mol of methane react with 1.094 mol of chlorine and produced 1.094 mol of chloromethane and 1.094 mol of HCl.

Hence,

Number of moles of chloromethane = 1.094mol

Number of moles of HCl = 1.094mol.

Step (3) Calculation for volume of products.

Volume of chloromethane = (1.094mol)(22.4L/mol)

volume of Chloromethane = 24.5L

volume of HCl = (1.094mol)(22.4L/mol)

Volume of HCl = 24.5L

step(4) Calculation for total volume of products

total volume of products = volume of Chloromethane + volume of HCl

Total volume of Products = 22.5L + 22.5L

Total volume of Products = 45.0L