Question

10 ml of 1M Bacl2 solution and 5ml 0.5 M K2SO4 are mixed together to precipitate out BaS04 .The amount of BaSO4 precipited will be

a) 0.005.mol b) 0.00025mol

c)0.025mol. c)0.0025 mol

Answer 1

Volume in litres x molarity = number of moles
Therefore,
Number of moles of BaCl2 = 0.01 x 1 = 0.01
Number of moles of K2(SO4) = 0.005 x 0.5 = 0.0025
Reaction between BaCl2 and K2(SO4) :
BaCl2 + k2(SO4) →BaSO4 + 2kCl
According to the equation:
1 moles of BaCl2 give 1 moles of BaSO4
0.0025 moles of BaCl2 will give 0.0025 moles of BaSO4

Now,
Molar mass of BaSO4 = 137g + 32g + 4 x 16 g
= 233 g/mol
Mass of 0.0025 moles of BaSO4 = 0.0025 x 233
= 0.5825 g

The answer is 0.0025 mole