Question

A sample contains some potassium sulfate K2SO4 and weighs 2.000g, after reaction with barium it yielded a precipitate of BaSO4 weighing 2.500g.calculate the percent K2SO4 in the original sample. Thanks

Answer 1

Molar mass of BaSO4,

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

= 1*137.3 + 1*32 + 4*16

= 233.3 g/mol

mass(BaSO4)= 2.500 g

number of mol of BaSO4,

n = mass/molar mass

=(2.5 g)/(233.3 g/mol)

= 0.01072 mol

The reaction taking place is:

K2SO4 + Ba —> BaSO4 + 2 K

from reaction,

mol of K2SO4 actually reacted = number of mol of BaSO4

mol of K2SO4 actually reacted = 0.01072 mol

Molar mass of K2SO4,

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

= 2*39 + 1*32 + 4*16

= 174 g/mol

mass of K2SO4,

m = number of mol * molar mass

= 0.01072 mol*174 g/mol

= 1.865 g

% K2SO4 = mass of K2SO4 calculated * 100 / mass of K2SO4 present

% K2SO4 = 1.865 * 100 / 2.000

% K2SO4 = 93.25 %

Answer: 93.25 %​

Molar mass of BaSO4,

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

= 1*137.3 + 1*32 + 4*16

= 233.3 g/mol

mass(BaSO4)= 2.500 g

number of mol of BaSO4,

n = mass/molar mass

=(2.5 g)/(233.3 g/mol)

= 0.01072 mol

The reaction taking place is:

K2SO4 + Ba —> BaSO4 + 2 K

from reaction,

mol of K2SO4 actually reacted = number of mol of BaSO4

mol of K2SO4 actually reacted = 0.01072 mol

Molar mass of K2SO4,

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

= 2*39 + 1*32 + 4*16

= 174 g/mol

mass of K2SO4,

m = number of mol * molar mass

= 0.01072 mol*174 g/mol

= 1.865 g

% K2SO4 = mass of K2SO4 calculated * 100 / mass of K2SO4 present

% K2SO4 = 1.865 * 100 / 2.000

% K2SO4 = 93.25 %

Answer: 93.25 %