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
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.
Thank you so much, now I understand ... I think :)
To find the mass of silver chromate (Ag2CrO4) produced, we need to determine the limiting reactant and then use stoichiometry to calculate the mass. Let's break it down step by step:
1. Start by finding the number of moles of silver nitrate (AgNO3) and potassium chromate (K2CrO4):
Moles of AgNO3 = Volume (L) × Molarity (mol/L) = 0.120 L × 0.320 mol/L = 0.0384 mol
Moles of K2CrO4 = Volume (L) × Molarity (mol/L) = 0.040 L × 0.320 mol/L = 0.0128 mol
2. Next, we need to determine which reactant is the limiting reactant, meaning the one that is completely consumed first. To do this, we compare the moles of each reactant based on their stoichiometric ratios.
The balanced equation for the reaction is:
2AgNO3 + K2CrO4 -> Ag2CrO4 + 2KNO3
From the equation, we can see that the stoichiometric ratio between AgNO3 and Ag2CrO4 is 2:1. Therefore, 2 moles of AgNO3 would produce 1 mole of Ag2CrO4.
Moles of Ag2CrO4 that can be produced from AgNO3 = 0.0384 mol / 2 = 0.0192 mol
Since the stoichiometric ratio is 2:1, the moles of Ag2CrO4 that can be produced from K2CrO4 would be 0.0128 mol × (1 mol Ag2CrO4 / 2 mol K2CrO4) = 0.0064 mol
Comparing the moles of Ag2CrO4 that can be produced from each reactant, we see that AgNO3 is the limiting reactant because it produces fewer moles of Ag2CrO4.
3. Now, we can determine the moles of Ag2CrO4 produced from the limiting reactant.
Moles of Ag2CrO4 = moles of limiting reactant = 0.0192 mol
4. Finally, we need to calculate the mass of Ag2CrO4 using its molar mass.
The molar mass of Ag2CrO4 = (2 × atomic mass of Ag) + atomic mass of Cr + 4 × atomic mass of O
= (2 × 107.87 g/mol) + 52.00 g/mol + (4 × 16.00 g/mol)
≈ 331.87 g/mol
Mass of Ag2CrO4 = Moles × Molar mass
= 0.0192 mol × 331.87 g/mol
≈ 6.37 g
Therefore, approximately 6.37 grams of silver chromate (Ag2CrO4) would be produced when 120 mL of 0.320 M silver nitrate (AgNO3) is reacted with 40 mL of 0.320 M potassium chromate (K2CrO4).