Posted by Some kid on .
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...

chemistry 
DrBob222,
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. 
chemistry 
Some kid,
i got an answer of 3.27 g, i hope that's right

chemistry 
DrBob222,
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. 
chemistry 
DrBob222,
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. 
chemistry 
Jenn,
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.