Question

What mass of silver chromate is produced from 1.23 mol of silver nitrate

Answer 1

What mass of silver chromate is produced from 1.23 mol of silver nitrate.

Solution:

The chemical reaction of silver chromate formation is following:

2AgNO₃ + K₂CrO₄ = Ag₂CrO₄ (s) + 2KNO₃

We can calculate the mass of silver nitrate using ration:

m(AgNO₃) = n*M(AgNO₃) where n is number of moles; M is molar mass in g/mol.

Thus:

m(AgNO₃) = 1.23*(108 + 14 + 3*16) = 209.1 (g)

Using the chemical reaction we can determine the mass of silver chromate:

209.1 g X g

2AgNO₃ + K₂CrO₄ = Ag₂CrO₄ (s) + 2KNO₃

2*170 332

Make a proportion:

209.1/340 = X/332

where:

X = m (Ag₂CrO₄) = 209.1*332/340 = 204.18 (g)

Answer: m (Ag₂CrO₄) = 204.18 g.