## Check significant figures redox titration calculation

**Anonymous**

Q1.What mass FeSO[sub]4[/sub],(NH[sub]4[/sub])[sub]2[/sub]SO[sub]4[/sub],6H[sub]2[/sub]O you will need to prepare 100mL of a 0.05M solution.

c=n/v

n=cxv

n(above compound)=.05 x .1000

=.00500 mol

Mr(above compound) =55.9+32.1+16.0x4+2x(14.0+1.0x4)+32.1+16.0x4+6x(16.0+1.0x2)

=392.1g/mol

n=m/Mr

m=Mrxn

m(above compound)=392.1x.00500

=1.96 g

Q2. How many moles of Fe[sup]2[/sup][sup]+[/sup] are contained in 25.0 ml of 0.05M solution.

c=nxv

n=cxv

n[Fe[sup]2[/sup][sup]+[/sup]]=0.05x0.0250

=1.25E[sup]-3[/sup]mol

Q.3 How many moles of MnO[sub]4[/sub][sup]-[/sup] will be required to react with the 25ml of 0.05M Fe[sup]2[/sup][sup]+[/sup] solution in Q2.

MnO[sub]4[/sub][sup]-[/sup]+8H[sup]+[/sup]+5Fe[sup]2[/sup][sup]+[/sup]-->Mn[sup]2[/sup][sup]+[/sup]+5Fe[sup]3[/sup][sup]+[/sup]+4H[sub]2[/sub]O

n[MnO[sub]4[/sub][sup]-[/sup]]=1.25E[sup]-3[/sup]x1/5

=2.50E[sup]-4[/sup]mol

Anonymous| 20 February 2005 - 9:31am## Re: Check significant figures redox titration calculation

[quote="avogadro1"]Q2. How many moles of Fe[sup]2[/sup][sup]+[/sup] are contained in 25.0 ml of 0.05M solution.

[b][color=red]c=nxv[/color][/b]

n=cxv

n[Fe[sup]2[/sup][sup]+[/sup]]=0.05x0.0250

=[b]1.25E[sup]-3[/sup][/b]mol[/quote]

Yup, that all looks fine (although I haven't checked the calculated results; I'm eating my breakfast at the moment).

I just thought I should point out two minor things.

The bold bit in red is incorrect; c=n/v

However, you correctly used n=cxv, so I assume that was just a typo.

As for the other bit in bold, if you are going to write out the result in a textbook, or anything other than Excel, you want to replace the E[sup]-3[/sup] with x10[sup]-3[/sup].

So 1.25E[sup]-3[/sup] == 1.25x10[sup]-3[/sup] == 1.25mMol (milli Moles)

Anonymous| 20 February 2005 - 8:50pmI see c=n/v

and 1.25x10[sup]-3[/sup]