Question

Consider the precipitation reaction below:

Ca2+ + SO4^2- -> CaSO4

a) balance this reaction (if necessary)

b) predict how many g of CaSO4 would be formed if you add 1 g of Ca2+ and 1 g of SO4^2- to a solution,

c) identify the limiting reactant.

Answer 1

1.

reaction is alredy balanced

Ca2+ + SO4^2- -----> CaSO4

2.

moles of Ca2+ = 1 g / 40.077 g/mol = 0.025 mol

moles of SO₄^2- = 1 g/ 96.064 g/mol = 0.0104 mol

from the balanced equation it is very clear that

1 mol of Ca2+ react with 1 mol of SO4- to give 1 mol of CaSO4 accordingly

0.025 mol of Ca2+ need 0.025 mol of SO42- but we have only 0.0104 mol of

so limiting agent is SO42-

0.0104 mol of SO42- react with 0.0104 mol of Ca2+ and form the 0.0104 mol of CaSO4

mass of CaSO4 = 0.0104 mol x 136.1406 g/mol = 1.42 g