2
answers
0
watching
412
views

Calculate the amount of manganese (54.94 g/mol) from a processed ore using the following data.

Standardization data: 

2.000 mL KMnO4 = 0.02015 g Na2C2O4 (134 g/mol) 

Sample analysis data: 

Weight of sample = 0.1012 g 

Na2C2O4 used(in excess)= 0.3200 g 

KMnO4 used for back-titration = 9.93 mL 

Rxn: 2MnO4^- + 5C2O42^- + 16H^+ → 2Mn2^+ + 10CO2 + 8H2O

For unlimited access to Homework Help, a Homework+ subscription is required.

Avatar image
Liked by mohamar135 and 3 others

Unlock all answers

Get 1 free homework help answer.
Already have an account? Log in
Avatar image
Read by 2 people
Already have an account? Log in
discord banner image
Join us on Discord
Chemistry Study Group
Join now

Related textbook solutions

Related questions

Weekly leaderboard

Start filling in the gaps now
Log in