Online problems and solutions in Chemistry

Calculate the number of grams of silver chromate formed from 25.0 mL of 0.125 M silver nitrate and 15.0 mL of 0.120 M potassium chromate.

Step 1_ Write and balance the reaction to see how many moles of silver nitrate could form from a mole of each reagent
2AgNO₃ + K₂CrO₄ → Ag₂CrO_{4 +} 2KNO₃
Each mole of K₂CrO₄ can react with 2 moles of AgNO₃ to produce 1 mole of Ag₂CrO₄.
Each mole of AgNO₃ can react with 0.5 moles of K₂CrO₄ to produce 0.5 mole of Ag₂CrO₄.
Step 2_ Calculate how many moles of each reactant you have.
(0.0250 L)(0.125 mol/L) = 0.003125 moles of AgNO₃
(0.0150 L)(0.120 mol/L) = 0.00180 moles of K₂CrO₄
Step 3_ Figure out how much product will form, and which reactant is the limiting reagent (or if they are present in equivalent amounts).
Since each mole of K₂CrO₄ can react with 2 moles of AgNO₃, 0.00180 moles of K₂CrO₄ could react with 0.00360 moles of AgNO₃. However you do not have that much AgNO₃. Silver nitrate is the limiting reagent, and K₂CrO₄ is in excess.
0.003125 moles of AgNO₃ will react with 0.00161 moles of K₂CrO₄ to produce 0.00161 moles of Ag₂CrO₄.
Step 4_ Find the molecular weight (or formula weight) of Ag₂CrO₄, and calculate the number of grams formed.
Your instructor probably expect you to find the fromula weight for Ag₂CrO₄ by adding atomic weights. I added up to 331.73. That would mean 1 mole has a mass of 331.73 g
(0.00161 moles)(331.73 g/mol) = 0.534 g Ag₂CrO4