A community for students.
Here's the question you clicked on:
 0 viewing
samigupta8
 3 years ago
how many milliliters of a 0.05 M kmno4 solution are required to oxidise 2 g of feso4 in a dilute acid solution
samigupta8
 3 years ago
how many milliliters of a 0.05 M kmno4 solution are required to oxidise 2 g of feso4 in a dilute acid solution

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well the answer is 52.57 mL of a .05 M KMnO4 solution, but let me get back to you with the explanation.

samigupta8
 3 years ago
Best ResponseYou've already chosen the best response.0bt plssss.... make it fast for i hve to ask 1 more ques..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ask the other question while im here, Molarity is always a tough subject to explain

samigupta8
 3 years ago
Best ResponseYou've already chosen the best response.0i'll ask it afterwrds firstly can u explain me how u got it

samigupta8
 3 years ago
Best ResponseYou've already chosen the best response.0plss....do tell me fast

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0first you find out how many moles there are in 2 grams of FeSO4. (2grams/molarmass) >( 2g/151.908g/mol) = .0132 moles of FeSO4 >now you divide .0132 moles of FeS04 by 5 because of the stoichiometric ratio between Fe2+ and Mn2+. MnO4 + 8H+ + 5Fe2+ > Mn2+ + 4H2O + 5Fe3+ thats the balanced equation. There's a coefficient of 5 on Fe2+ therefore like i said before we divide our .0132moles of FeS04 by 5 so that it can give us an equal ratio to Mn2+ > .0132moles/5 = .00264 moles of FeS04 The question asks us how many mL's of .05 Molar solution of KMnO4 are required to oxidize 2 grams of FeSO4. So far we converted the grams into moles, and in the end we got .00264 moles of FeSO4. Now we need the moles of KMnO4. Luckily they gave us its Molarity. > Molarity is (# of moles solute/1 L). Our molarity is .05 and anything divided by 1(Liter) is still the same number therefore we can conclude that the number of moles of KMn04 is .05. We want mL's here so we are going to divide the number of FeSO4 moles by KMnO4 moles. > .00264 moles / .05 moles =.0528 This .0528 has no units since the moles cancelled out in the last calculation therefore we can multiply it by either 1L/or 1000mL. In our case our answer is looking for mL so we multiply > .0528 by 1000mL = 52.8 mL of .05 M KMnO4 is used to oxidize 2 grams of FeSO4. If you have questions on the stoichiometric stuff i posted a link below: http://www.chembuddy.com/?left=balancingstoichiometry&right=halfreactionsmethod hope that helped
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.