In: Chemistry
For the following reaction, 2.45 grams of methane (CH4) are allowed to react with 26.8 grams of carbon tetrachloride.
methane (CH4)(g) + carbon tetrachloride(g) ---> dichloromethane (CH2Cl2)(g)
What is the maximum mass of dichloromethane (CH2Cl2) that can be formed?
What is the formula for the limiting reagent?
What mass of the excess reagent remains after the reaction is complete? in grams
1)
Molar mass of CH4,
MM = 1*MM(C) + 4*MM(H)
= 1*12.01 + 4*1.008
= 16.042 g/mol
mass(CH4)= 2.45 g
number of mol of CH4,
n = mass of CH4/molar mass of CH4
=(2.45 g)/(16.042 g/mol)
= 0.1527 mol
Molar mass of CCl4,
MM = 1*MM(C) + 4*MM(Cl)
= 1*12.01 + 4*35.45
= 153.81 g/mol
mass(CCl4)= 26.8 g
number of mol of CCl4,
n = mass of CCl4/molar mass of CCl4
=(26.8 g)/(153.81 g/mol)
= 0.1742 mol
Balanced chemical equation is:
CH4 + CCl4 ---> 2 CH2Cl2
1 mol of CH4 reacts with 1 mol of CCl4
for 0.152724 mol of CH4, 0.152724 mol of CCl4 is required
But we have 0.174241 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)*0.152724
= 0.305448 mol
mass of CH2Cl2 = number of mol * molar mass
= 0.3054*84.93
= 25.94 g
Answer: 25.9 g
2)
CH4 is limiting reagent
3)
According to balanced equation
mol of CCl4 reacted = (1/1)* moles of CH4
= (1/1)*0.152724
= 0.152724 mol
mol of CCl4 remaining = mol initially present - mol reacted
mol of CCl4 remaining = 0.174241 - 0.152724
mol of CCl4 remaining = 0.021517 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
= 2.152*10^-2 mol * 153.81 g/mol
= 3.31 g
Answer: 3.31 g