mols KMnO4 = grams/molar mass = approx 0.033 but you need to this and all of the calculations that follow more accurately.
mols ethanol = approx 0.05
Convert mols KMnO4 to mols H2SO4. That's 0.033 x 6/4 = 0.05
Convert mols ethanol to mols H2SO4. That's 0.05 x 6/5 = 0.06
The limiting reagent is KMnO4 and you will use approx 0.05 mols H2SO4
Then convert mols H2SO4 to grams. g = mols x molar mass.
Check my work.
A chemist mixes 5.216 g of potassium permanganate, 2.318 g of ethanol, and excess sulfuric acid. These chemicals react as follows:
4 KMnO4 + 5 C2H5OH + 6 H2SO4 ---> 4 MnSO4 + 5 CH3COOH + 2 K2SO4 + 11 H20
What mass of sulfuric acid is consumed in this reaction, assuming that the reaction goes to completion?
3 answers
I got 4.854 g of H2SO4. Thank you so much.
I did it right the first time just has trouble understanding the problem in my head.
To find grams of excess reactant I just have to convert moles of ethanol to g right?
I did it right the first time just has trouble understanding the problem in my head.
To find grams of excess reactant I just have to convert moles of ethanol to g right?
excess mol ethanol x molar mass ethanol = grams ethanol in excess
1 answer