Question

For the following reaction, 86.7 grams of barium chloride are allowed to react with 68.5 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?  grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete?  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)= 86.7 g

use:

number of mol of BaCl2,

n = mass of BaCl2/molar mass of BaCl2

=(86.7 g)/(2.082*10^2 g/mol)

= 0.4164 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)= 68.5 g

use:

number of mol of K2SO4,

n = mass of K2SO4/molar mass of K2SO4

=(68.5 g)/(1.743*10^2 g/mol)

= 0.3931 mol

Balanced chemical equation is:

BaCl2 + K2SO4 ---> BaSO4 + 2 KCl

1 mol of BaCl2 reacts with 1 mol of K2SO4

for 0.4164 mol of BaCl2, 0.4164 mol of K2SO4 is required

But we have 0.3931 mol of K2SO4

so, K2SO4 is limiting reagent

we will use K2SO4 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 K2SO4

= (1/1)*0.3931

= 0.3931 mol

use:

mass of BaSO4 = number of mol * molar mass

= 0.3931*2.334*10^2

= 91.73 g

Answer: 91.7 g

2)

K2SO4 is limiting reagent

Answer: K2SO4

3)

According to balanced equation

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

= (1/1)*0.3931

= 0.3931 mol

mol of BaCl2 remaining = mol initially present - mol reacted

mol of BaCl2 remaining = 0.4164 - 0.3931

mol of BaCl2 remaining = 2.336*10^-2 mol

Molar mass of BaCl2,

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

= 1*137.3 + 2*35.45

= 208.2 g/mol

use:

mass of BaCl2,

m = number of mol * molar mass

= 2.336*10^-2 mol * 2.082*10^2 g/mol

= 4.863 g

Answer: 4.86 g