Question

When 1.30 x 10² mL of 0.0100 M BaCl₂ is mixed with 1.50 x 10²mL of 0.00500 M H₂SO₄, what mass of precipitate is formed ?

show working out

Answer 1

volume , V = 130 mL

= 0.13 L

number of mol,

n = Molarity * Volume

= 0.01*0.13

= 1.3*10^-3 mol

volume , V = 150 mL

= 0.15 L

number of mol,

n = Molarity * Volume

= 0.005*0.15

= 7.5*10^-4 mol

Balanced chemical equation is:

BaCl2 + H2SO4 ---> BaSO4 + 2 HCl

1 mol of BaCl2 reacts with 1 mol of H2SO4

for 0.0013 mol of BaCl2, 0.0013 mol of H2SO4 is required

But we have 0.0008 mol of H2SO4

so, H2SO4 is limiting reagent

we will use H2SO4 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 H2SO4

= (1/1)*7.5*10^-4 mol

= 7.5*10^-4 mol

mass of BaSO4 = number of mol * molar mass

= 7.5*10^-4*2.334*10^2

= 0.175 g

Answer: 0.175 g