Question

How many grams of BaSO4 are formed when a 250 mL of .800 M of sodium sulfate solution is added to 300.0 mL of aqueous 0.455 M BaCl2?(Hint: this is also a limiting reactant problem)

Answer 1

volume of Na2SO4, V = 250 mL
= 0.25 L

number of mol of Na2SO4,
n = Molarity * Volume
= 0.8*0.25
= 0.2 mol
volume OF BaCl2 , V = 300.0 mL
= 0.3 L


number of mol of BaCl2,
n = Molarity * Volume
= 0.455*0.3
= 0.1365 mol
Balanced chemical equation is:
Na2SO4 + BaCl2 ---> BaSO4 + 2 NaCl


1 mol of Na2SO4 reacts with 1 mol of BaCl2
for 0.2 mol of Na2SO4, 0.2 mol of BaCl2 is required
But we have 0.1365 mol of BaCl2

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)*0.1365
= 0.1365 mol


mass of BaSO4 = number of mol * molar mass
= 0.1365*2.334*10^2
= 31.9 g
Answer: 31.9 g