What mass of silver chromate (in grams) will be produced when 100.0 mL of 2.30 M silver nitrate is mixed with 300.0 mL of 1.60 M sodium chromate?

38.15 g of silver chromate

Explanation:

The equation of the reaction is;

2AgNO3(aq)+Na2CrO4(aq)→Ag2CrO4(s)↓+2NaNO3(aq)

number of  moles of silver nitrate = 100/1000 * 2.3 = 0.23 moles

number of moles of sodium chromate = 300/1000 * 1.6 = 0.48 moles

We shall now obtain the limiting reactant:

From the reaction equation,

2 moles of silver nitrate yields 1 mole of silver chromate

0.23 moles of silver nitrate will yield 0.23 * 1/2 = 0.115 moles of silver chromate

for sodium chromate:

1 mole of sodium chromate yields 1 mole of silver chromate

0.48 moles of sodium chromate yields 0.48 moles of silver chromate

Hence silver nitrate is the limiting reactant (it produced a lower number of moles of silver chromate-the product).

Hence, mass of silver chromate = 331.73 g/mol * 0.115 moles = 38.15 g of silver chromate

the answer is c i believe