mix 120ml of 0.320M silver nitrate with 40ml of 0.320M potasium chromate. What mass of silver chromate is produced?
this is what i have so far
2AgNO3 + K2CrO4 --> Ag2CrO4 + 2KNO3
molar mass of 2agno3 = 277.81
molar mass of k2cro4 = 194.2
molar mass of ag2cro4 = 331.8
limiting reactant
2agno3 .120*.320= .0384
k2cro4 .040*.320= .0128
limiting reactant would be k2cro4?
what are my next steps?
would it be .0128*331.8 = 4.25 ?
i am so lost!
I answered your original post in detail. I didn't understand most of what you posted as your response to your first post but this one looks ok. You can go back there to compare if you wish. Where are you so lost? This looks ok to me. The only problem I see, if you were working this for me on an exam, is that you didn't show why K2CrO4 is the limiting reagent.
lol to be honest I have no idea why it is the limiting reactant.
Thank you so much for posting it helped a great deal when compared to the previous one.
To solve this problem, you're on the right track by identifying the limiting reactant. The limiting reactant is the one that is completely consumed and determines the amount of product that can be formed.
To find the limiting reactant, you correctly calculated the moles of each reactant using their concentrations and volumes.
For Silver Nitrate (AgNO3):
Moles of AgNO3 = 0.120 L * 0.320 M = 0.0384 mol
For Potassium Chromate (K2CrO4):
Moles of K2CrO4 = 0.040 L * 0.320 M = 0.0128 mol
Comparing the moles of each reactant, you correctly concluded that Potassium Chromate (K2CrO4) is the limiting reactant because it has fewer moles than Silver Nitrate (AgNO3).
Now, you need to use stoichiometry to find the mass of Silver Chromate (Ag2CrO4) produced.
From the balanced equation, you know that 1 mole of Silver Chromate (Ag2CrO4) is produced for every 2 moles of Potassium Chromate (K2CrO4).
Now, multiply the moles of Potassium Chromate (K2CrO4) by the mole ratio:
Moles of Ag2CrO4 = 0.0128 mol * (1 mol Ag2CrO4 / 2 mol K2CrO4) = 0.0064 mol
Next, use the molar mass of Silver Chromate (Ag2CrO4) to convert moles to grams:
Mass of Ag2CrO4 = 0.0064 mol * 331.8 g/mol = 2.1152 g
So, the mass of Silver Chromate (Ag2CrO4) produced is approximately 2.12 grams.
It's important to note that your calculation was correct, but you made an arithmetic error in your final step. Instead of 4.25, the correct result is 2.1152 g.
I hope this explanation helps clarify the steps for solving this problem.