How many milliliters of .238 M KMnO4 are needed to react with 3.36 g of iron(II) sulfate, FeSO4? The reaction is as follows: 10FeSO4 (aq) + 2KMnO4 (aq) + 8H2SO4 --> (the products don't matter)
I got 465mL, but I'm not sure if that's right. I think my main concern is where the 10:2 ratio (5:1) comes into play. I have: .238M x __V = .022118709(5), but I don't know if that's where the 5 belongs... I have a huge chem test tomorrow, thanks for the help!
PS. I know to divide by 1000 to turn whatever that number comes out to be into mL
I got 465mL, but I'm not sure if that's right. I think my main concern is where the 10:2 ratio (5:1) comes into play. I have: .238M x __V = .022118709(5), but I don't know if that's where the 5 belongs... I have a huge chem test tomorrow, thanks for the help!
PS. I know to divide by 1000 to turn whatever that number comes out to be into mL
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molarity = mol/L
stoichiometry from grams FeSO4 to mol KMnO4
3.36g FeSO4 x (1 mol FeSO4/152 g FeSO4) x (2 mol KMnO4/10 mol FeSO4)
= 0.004421 mol KMnO4
plug into molarity formula.
0.238 M KMnO4 = 0.004421 mol KMnO4 / ? L
? = 0.0186 L
converting that to mL:
0.0186 L x (1 mL/10^-3 L) = 18.6 mL
hope that helps! good luck on your exam.
stoichiometry from grams FeSO4 to mol KMnO4
3.36g FeSO4 x (1 mol FeSO4/152 g FeSO4) x (2 mol KMnO4/10 mol FeSO4)
= 0.004421 mol KMnO4
plug into molarity formula.
0.238 M KMnO4 = 0.004421 mol KMnO4 / ? L
? = 0.0186 L
converting that to mL:
0.0186 L x (1 mL/10^-3 L) = 18.6 mL
hope that helps! good luck on your exam.