Question

What mass of precipitate is formed when 135.00ml .25 M calcium chloride is reacted with 145.00 ml of .55M potassium sulfate? Answer: 4.6 g

Answer 1

volume of CaCl2, V = 135.00 mL

= 0.135 L

we have below equation to be used:

number of mol in CaCl2,

n = Molarity * Volume

= 0.25*0.135

= 3.375*10^-2 mol

volume of K2SO4, V = 145.00 mL

= 0.145 L

we have below equation to be used:

number of mol in K2SO4,

n = Molarity * Volume

= 0.55*0.145

= 7.975*10^-2 mol

we have the Balanced chemical equation as:

CaCl2 + K2SO4 ---> CaSO4 + 2 KCl

1 mol of CaCl2 reacts with 1 mol of K2SO4

for 3.375*10^-2 mol of CaCl2, 3.375*10^-2 mol of K2SO4 is required

But we have 7.975*10^-2 mol of K2SO4

so, CaCl2 is limiting reagent

we will use CaCl2 in further calculation

Molar mass of CaSO4 = 1*MM(Ca) + 1*MM(S) + 4*MM(O)

= 1*40.08 + 1*32.07 + 4*16.0

= 136.15 g/mol

From balanced chemical reaction, we see that

when 1 mol of CaCl2 reacts, 1 mol of CaSO4 is formed

mol of CaSO4 formed = (1/1)* moles of CaCl2

= (1/1)*3.375*10^-2

= 3.375*10^-2 mol

we have below equation to be used:

mass of CaSO4 = number of mol * molar mass

= 3.375*10^-2*1.362*10^2

= 4.6 g

Answer: 4.6 g