Question

For the following reaction, 79.5 grams of barium chloride are allowed to react with 71.2 grams of potassium sulfate.

barium chloride(aq) + potassium sulfate(aq) -> barium sulfate(s) + potassium chloride(aq)

What is the maximum amount of barium sulfate 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 BaCl2,

MM = 1*MM(Ba) + 2*MM(Cl)

= 1*137.3 + 2*35.45

= 208.2 g/mol

mass(BaCl2)= 79.5 g

use:

number of mol of BaCl2,

n = mass of BaCl2/molar mass of BaCl2

=(79.5 g)/(2.082*10^2 g/mol)

= 0.3818 mol

Molar mass of K2SO4,

MM = 2*MM(K) + 1*MM(S) + 4*MM(O)

= 2*39.1 + 1*32.07 + 4*16.0

= 174.27 g/mol

mass(K2SO4)= 71.2 g

use:

number of mol of K2SO4,

n = mass of K2SO4/molar mass of K2SO4

=(71.2 g)/(1.743*10^2 g/mol)

= 0.4086 mol

Balanced chemical equation is:

BaCl2 + K2SO4 ---> BaSO4 + 2 KCl

1 mol of BaCl2 reacts with 1 mol of K2SO4

for 0.3818 mol of BaCl2, 0.3818 mol of K2SO4 is required

But we have 0.4086 mol of K2SO4

so, BaCl2 is limiting reagent

we will use BaCl2 in further calculation

Molar mass of BaSO4,

MM = 1*MM(Ba) + 1*MM(S) + 4*MM(O)

= 1*137.3 + 1*32.07 + 4*16.0

= 233.37 g/mol

According to balanced equation

mol of BaSO4 formed = (1/1)* moles of BaCl2

= (1/1)*0.3818

= 0.3818 mol

use:

mass of BaSO4 = number of mol * molar mass

= 0.3818*2.334*10^2

= 89.11 g

Answer: 89.1 g

2)

BaCl2 is limiting reagent

3)

According to balanced equation

mol of K2SO4 reacted = (1/1)* moles of BaCl2

= (1/1)*0.3818

= 0.3818 mol

mol of K2SO4 remaining = mol initially present - mol reacted

mol of K2SO4 remaining = 0.4086 - 0.3818

mol of K2SO4 remaining = 2.672*10^-2 mol

use:

mass of K2SO4,

m = number of mol * molar mass

= 2.672*10^-2 mol * 1.743*10^2 g/mol

= 4.656 g

Answer: 4.66 g