Question

How many mL of 0.261 M Na2SO4 are needed to react with 27.5 mL of 0.189 M BaCl2 solution? Na2SO4(aq) + BaCl2(aq) ---> BaSO4(s) + 2NaCl(aq)

Answer 1

Given that

Na2SO4 (aq )+ BaCl2 (aq) ----- > BaSO4 (s) + 2 NaCl (aq)

Na2SO4 = 0.261 M

BaCl2 = 27.5 ml of 0.189 M

Number of moles = molarity * volume in L

= 0.189* 27.5/1000

= 0.0052 moles BaCl2

Now calculate the moles of Na2SO4 as follows:

0.0052 moles BaCl2 *1 mole Na2SO4/1 moles BaCl2

=0.0052 moles Na2SO4

Molarity = number of moles / volume in L

Volume in L = Number of moles / molarity

= 0.0052 moles Na2SO4/ 0.261 mole/ L

= 0.0199 L

= 0.020 L

0.020 L*1000 ml/1 L

= 20 ml