1. Write and balance the equation for oxalate and permananate, then calculate the M of the permanganate.
2. Write and balance the equation for the Sn and permanganate.
3. mols MnO4^- = M x L = ?
Convert mols MnO4^- to mols Sn using the coefficients in the balanced equation. grams Sn = mols Sn x atomic mass Sn.
4. %Sn = (mass Sn/mass sample)*100 = ?
Post your work if you have trouble and I can help you through it.
A 0.1818 g sample of sodium oxalate required 28.12 ml of a potassium permanganate solution to reach the endpoint (assume the reaction was carried out in excess acid). A 5.000 g sample of tin (II) sulfate containing potassium sulfate as an impurity was found to require 35.00 ml of the above permanganate solution (in excess acid) to reach the endpoint. What was the percentage of tin in the sample?
I don't have any clue how to do this, help?
2 answers
5CrO42-+2MnO4-+16H+->2MN2++8H2O+10CO2
.1818gC2O4*1mol/88.019gNaC2O4*2molMnO4-/5molC2O42-=8.2618^-4mol/.02812l=.02938M
(Was I supposed to use the molar mass of NaCrO4 or just CrO4 for this step?)
2)16H++5Sn2++2MnO4- -->5Sn4++2Mn2++8H2O
.02938M*.035L=.001028mol MnO4-
.001028molMnO4-*5molSn2+/2molMnO4-*118.71gSn=.3052gSn
.3052gSn/5gSn*100=6.104%
Thank you for your help.
.1818gC2O4*1mol/88.019gNaC2O4*2molMnO4-/5molC2O42-=8.2618^-4mol/.02812l=.02938M
(Was I supposed to use the molar mass of NaCrO4 or just CrO4 for this step?)
2)16H++5Sn2++2MnO4- -->5Sn4++2Mn2++8H2O
.02938M*.035L=.001028mol MnO4-
.001028molMnO4-*5molSn2+/2molMnO4-*118.71gSn=.3052gSn
.3052gSn/5gSn*100=6.104%
Thank you for your help.
1 answer