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.
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
4 answers
this is what I came up with
2AgNO3+k2CrO4--->Ag2CrO4+2KNO3
molar mass of 2agno3 = 277.18
molar mass of k2cro4 = 194.2
molar mass of ag2cro4 = 331.08
194.2*277.18/120ml=500ml
limiting reactant k2cro4
194.2g*331.08g/500=128.60
128.60*331.08/1000=42.6
is this anywhere near right?
2AgNO3+k2CrO4--->Ag2CrO4+2KNO3
molar mass of 2agno3 = 277.18
molar mass of k2cro4 = 194.2
molar mass of ag2cro4 = 331.08
194.2*277.18/120ml=500ml
limiting reactant k2cro4
194.2g*331.08g/500=128.60
128.60*331.08/1000=42.6
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.
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 :)
1 answer