Potassium permanganate reacts with sulfuric acid as per the equation shown below. Indicate the excess reactant after the reaction of 7.60 g KMnO4 and 10.32 g H2SO4.
4 KMnO4 (aq) + 6H2SO4 (aq) + 2K2SO4 (aq) + 4MnSO4 (aq)+ 6H2) (l) + 5O2 (g)
I have no idea where to start or what to do on this problem...
Look at the problem I did for you below (SrH2 + H2O problem).
This one is done the same way.
Convert KMnO4 to moles
Convert H2SO4 to moles.
Convert each one of the moles to moles of any product you wish. Then take the smaller one. That will be the limiting reagent. The other one will be the excess reagent. If you want to determine how much of it remains unreacted, calulate the amount used by the limiting reagent and subtract from the amount you started with.
Post your work if you get stuck and we can help you through it.
i got an answer of 3.27 g, i hope that's right
I may have made an error but I went through it in my head and didn't come up with your number. Two things you need to do.
First, I notice that the coefficients in the equation are double what they need to be. Second, there is no arrow to separate the reactants from the product. Please show where the arrow goes. Third, check you value. I'll do it by paper when I finish lunch.
ok.
First, the equation is correct. I didn't count the atoms correctly.
Second, I worked the problem on paper and came up with your answer. You probably worked it correctly. Good work for not knowing where to start.
I need help. I don't understand this one at all. My problem is the same, just the numbers are a little different.
Potassium permanganate reacts with sulfuric acid as per the equation shown below. Indicate the excess reactant after the reaction of 7.90 g KMnO4 and 10.32 g H2SO4.
To determine the excess reactant in a chemical reaction, you need to compare the moles of the limiting reactant (the reactant that is completely consumed) to the moles of the other reactant.
To solve this problem, you will follow these steps:
Step 1: Calculate the number of moles of each reactant.
Step 2: Determine the ratio of moles between the two reactants based on the coefficients of the balanced equation.
Step 3: Identify the limiting reactant.
Step 4: Calculate the number of moles of the product(s) formed using the stoichiometry of the balanced equation.
Step 5: Compare the moles of the limiting reactant to the moles of the other reactant to identify the excess reactant.
Let's go through the steps:
Step 1: Calculate the number of moles of each reactant.
To calculate the number of moles, you can use the formula: moles = mass / molar mass.
The molar mass of KMnO4 (potassium permanganate) is:
K: 39.10 g/mol
Mn: 54.94 g/mol
O: 16.00 g/mol
4 oxygen atoms x 16.00 g/mol = 64.00 g/mol
So, the molar mass of KMnO4 is: 39.10 + 54.94 + 64.00 = 158.04 g/mol.
Now, you can calculate the number of moles of KMnO4:
Moles of KMnO4 = mass / molar mass = 7.60 g / 158.04 g/mol = 0.0481 mol.
Similarly, calculate the number of moles of H2SO4 using its molar mass.
Step 2: Determine the ratio of moles between the two reactants.
Based on the balanced equation, the ratio of moles between KMnO4 and H2SO4 is 4:6 (from their coefficients), which simplifies to 2:3.
Step 3: Identify the limiting reactant.
To find the limiting reactant, compare the moles of the reactants. The reactant with fewer moles will be the limiting reactant.
In this case, since the ratio between KMnO4 and H2SO4 is 2:3, you have:
(0.0481 mol KMnO4) / (2 mol KMnO4 / 3 mol H2SO4) = 0.0721 mol H2SO4
Since you have 0.0721 mol of H2SO4, which is more than the actual moles of H2SO4 (calculated in Step 1), H2SO4 is the limiting reactant.
Step 4: Calculate the number of moles of the product(s) formed.
Based on the coefficients of the balanced equation, the molar ratio of H2SO4 to K2SO4 is 6:2, which simplifies to 3:1.
So, the moles of K2SO4 formed will be (0.0481 mol H2SO4) x (2 mol K2SO4 / 3 mol H2SO4) = 0.0321 mol K2SO4.
Step 5: Compare the moles of the limiting reactant to the moles of the other reactant.
The limiting reactant is H2SO4, and its moles are 0.0721 mol (calculated in Step 1).
To determine the excess reactant, subtract the moles of the limiting reactant from the moles of the other reactant:
Excess moles of KMnO4 = (0.0481 mol KMnO4) - (0.0721 mol H2SO4) = -0.024 mol.
Here, the negative value indicates that KMnO4 is in excess. However, since you cannot have a negative number of moles, the excess moles of KMnO4 should be considered as zero.
Therefore, in this reaction, KMnO4 is the excess reactant.