Question

For the following reaction, 2.44 grams of methane (CH₄) are allowed to react with 26.7 grams of carbon tetrachloride.

methane (CH₄) (g) + carbon tetrachloride (g) ---> dichloromethane (CH₂Cl₂) (g)

What is the maximum amount of dichloromethane (CH₂Cl₂) that can be formed? in grams

What is the formula for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? in grams

Answer 1

1)

Molar mass of CH4,

MM = 1*MM(C) + 4*MM(H)

= 1*12.01 + 4*1.008

= 16.042 g/mol

mass(CH4)= 2.44 g

number of mol of CH4,

n = mass of CH4/molar mass of CH4

=(2.44 g)/(16.042 g/mol)

= 0.1521 mol

Molar mass of CCl4,

MM = 1*MM(C) + 4*MM(Cl)

= 1*12.01 + 4*35.45

= 153.81 g/mol

mass(CCl4)= 26.7 g

number of mol of CCl4,

n = mass of CCl4/molar mass of CCl4

=(26.7 g)/(153.81 g/mol)

= 0.1736 mol

Balanced chemical equation is:

CH4 + CCl4 ---> 2 CH2Cl2

1 mol of CH4 reacts with 1 mol of CCl4

for 0.152101 mol of CH4, 0.152101 mol of CCl4 is required

But we have 0.173591 mol of CCl4

so, CH4 is limiting reagent

we will use CH4 in further calculation

Molar mass of CH2Cl2,

MM = 1*MM(C) + 2*MM(H) + 2*MM(Cl)

= 1*12.01 + 2*1.008 + 2*35.45

= 84.926 g/mol

According to balanced equation

mol of CH2Cl2 formed = (2/1)* moles of CH4

= (2/1)*0.152101

= 0.304201 mol

mass of CH2Cl2 = number of mol * molar mass

= 0.3042*84.93

= 25.83 g

Answer: 25.8 g

2)

CH4 is limiting reagent

3)

According to balanced equation

mol of CCl4 reacted = (1/1)* moles of CH4

= (1/1)*0.152101

= 0.152101 mol

mol of CCl4 remaining = mol initially present - mol reacted

mol of CCl4 remaining = 0.173591 - 0.152101

mol of CCl4 remaining = 0.02149 mol

Molar mass of CCl4,

MM = 1*MM(C) + 4*MM(Cl)

= 1*12.01 + 4*35.45

= 153.81 g/mol

mass of CCl4,

m = number of mol * molar mass

= 2.149*10^-2 mol * 153.81 g/mol

= 3.305 g

Answer: 3.31 g