posted by Krissy on .
mix 120ml of 0.320M silver nitrate with 40ml of 0.320M potasium chromate. What mass of silver chromate is produced?
I am stuck please help
Write the balanced reaction and determine the limiting reactant. That will tell you how many moles of silver chromate are produced. Convert that to grams using the molar mass.
Hint: Potassium chromate is K2CrO4
CrO4(-2) is divalent.
this is what I came up with
molar mass of 2agno3 = 277.18
molar mass of k2cro4 = 194.2
molar mass of ag2cro4 = 331.08
limiting reactant k2cro4
is this anywhere near right?
Your answer is close to a factor of 10 off but I don't understand most of what you did.
Here is how I would do it, following DrWLS' response.
moles AgNO3 = M x L = 0.32 x 0.120 = 0.0384 moles.
moles K2CrO4 = M x L = 0.32 x 0.040 = 0.0128 moles.
How much AgNO3 must we have if K2CrO4 is the limiting reagent? We must have 0.0128 x 2 = 0.0256 and we have that much; therefore, K2CrO4 is the limiting reagent. (We can check it by asking how much K2CrO4 we must have if AgNO3 is the limiting reagent. We must have 0.0384 x 1/2 = 0.0192 moles K2CrO4 and we don't have that much; therefore, K2CrO4 is the limiting reagent.)
So we will have 0.0128 moles Ag2CrO4 produced and that x molar mass 0.0128 x 331.73 = 4.246 grams Ag2CrO4 which rounds to 4.25 grams (I'm guessing we are allowed 3 significant figures although with 40 and 120 written as they are I can't tell). Check you work; perhaps you just made a decimal point error.
Thank you so much, now I understand ... I think :)