precipitation problem Did i do this right?

Question

When aqueous silver nitrate and sodium
chromate solutions are mixed, solid silver chromate forms in a
solution of sodium nitrate. If 257.8 mL of a 0.0468 M solution of
silver nitrate is added to 156.00 mL of a 0.0950 M solution of
sodium chromate, what mass of silver chromate (M = 331.8 g/mol)
will be formed?
This is a limiting-reactant problem because the amounts
of two reactants are given.
Strategy:
(1) Write the balanced equation.
I have AgNO3 + Na2CrO4 --> Ag2 CrO4 + NaNo3
(2) Calculate the number of moles of each reactant.
.2578L x .0468M = .01207mols AgNO3
.156L x .0950M = .01482mols Na2CrO4
(3) Determine the limiting reactant.
AgNO3
(4) Calculate the moles of product.
.01482/.01207 = 1.228mol
(5) Convert moles of product to mass of the product using molar mass
1.228mol x 331.8g/mol = 407.4g Is this right? It seems like a lot.

Answer 1

Your balanced equation should be:

2AgNO3 + Na2CrO4 ===> Ag2CrO4 + 2NaNO3

This may affect your calculation of the limiting reagent.