Question

When solutions containing 1.00 g of BaCl2 and 1.00 g of Na2S

Answer 1

) 1.00/208.23 = 0.00480 mol

c) 0.00480*233.43 = 1.12g

d) 1.00/142.04 = 0.00704

e) 0.00704*233.43 = 1.64g

f) BaCl2 is the limiting reagent,since it's availability is lower at 0.00480mol.

g) The theoretical yield is 1.12g.

h)1.04/1.12 * 100% = 92.9%

i) There will be following amount of NaSO4:

(0.00704-0.00480)*142.04 = 0.318g