Question

What is the mass of the precipitate that forms when 70 mL of 0.3 M BaCl2 reacts with 65 mL of 0.35 M Na2CO3? Express your answer to three significant figures and include the appropriate units.

Answer 1

volume , V = 70 mL

= 7*10^-2 L

number of mol,

n = Molarity * Volume

= 0.3*0.07

= 2.1*10^-2 mol

volume , V = 65 mL

= 6.5*10^-2 L

number of mol,

n = Molarity * Volume

= 0.35*0.065

= 2.275*10^-2 mol

Balanced chemical equation is:

BaCl2 + Na2CO3 ---> BaCO3 + 2 NaCl

1 mol of BaCl2 reacts with 1 mol of Na2CO3

for 0.021 mol of BaCl2, 0.021 mol of Na2CO3 is required

But we have 0.0227 mol of Na2CO3

so, BaCl2 is limiting reagent

we will use BaCl2 in further calculation

Molar mass of BaCO3,

MM = 1*MM(Ba) + 1*MM(C) + 3*MM(O)

= 1*137.3 + 1*12.01 + 3*16.0

= 197.31 g/mol

According to balanced equation

mol of BaCO3 formed = (1/1)* moles of BaCl2

= (1/1)*0.021

= 0.021 mol

mass of BaCO3 = number of mol * molar mass

= 2.1*10^-2*1.973*10^2

= 4.14 g

Answer: 4.14 g