# Help please, Moles question :)Watch

#1
Hey, I would love if someone, could point out where the
1 Mol MnO4- = 2.5 Mol NO2 In this question.

I have provided the Mark scheme answer below.
THanks

Q) In an experiment, all the sodium nitrite present in 1 kg of
sausage meat was extracted using distilled water and the solution
concentrated to 50cm3.
The solution was placed in a burette and titrated against 25.0cm3 of acidified 0.02 M potassium manganate(VII) solution.
10.5 cm3 of the solution was required to
reach the end point.

Calculate the percentage by mass of sodium nitrite present in the
sausage meat.

25 cm3 of 0.02 MnO4-
= 25 cm3 Ã— 10–3 Ã— 2 Ã— 10–2
= 5 Ã— 10–4 mol
1 mol MnO4 = 2.5 mol NO2
5 Ã— 10–4 mol MnO4 = 2.5 Ã— 5 Ã— 10–4 mol NO2
= 1.25 Ã— 10–3 mol
0.00125 mol NO2 are contained in 10.5 cm3 of solution
0.00125//10.5 Ã— 50 mol NO2 are contained in 50 cm3 of solution
= 0.00595 mol
NaNO2 = 23 + 14 + 32 = 69
0.00595 mol = 0.00595 Ã— 69 = 0.411 g
% = 0.411/1000 Ã— 100 = 0.0411%
0
8 years ago
#2
(Original post by ECullen)
Hey, I would love if someone, could point out where the
1 Mol MnO4- = 2.5 Mol NO2 In this question.

I have provided the Mark scheme answer below.
THanks

Q) In an experiment, all the sodium nitrite present in 1 kg of
sausage meat was extracted using distilled water and the solution
concentrated to 50cm3.
The solution was placed in a burette and titrated against 25.0cm3 of acidified 0.02 M potassium manganate(VII) solution.
10.5 cm3 of the solution was required to
reach the end point.

Calculate the percentage by mass of sodium nitrite present in the
sausage meat.

25 cm3 of 0.02 MnO4-
= 25 cm3 Ã— 10–3 Ã— 2 Ã— 10–2
= 5 Ã— 10–4 mol
1 mol MnO4 = 2.5 mol NO2
5 Ã— 10–4 mol MnO4 = 2.5 Ã— 5 Ã— 10–4 mol NO2
= 1.25 Ã— 10–3 mol
0.00125 mol NO2 are contained in 10.5 cm3 of solution
0.00125//10.5 Ã— 50 mol NO2 are contained in 50 cm3 of solution
= 0.00595 mol
NaNO2 = 23 + 14 + 32 = 69
0.00595 mol = 0.00595 Ã— 69 = 0.411 g
% = 0.411/1000 Ã— 100 = 0.0411%
The equivalence comes from the addition of balanced half equations:

manganate ions are oxidising agents:

MnO4- + 8H+ + 5e --> Mn2+ + 4H2O

Nitrite ions are reducing agents:

NO2- + H2O --> NO3- + 2e + 2H+

to add the two half equations together you must make the electrons equal by multiplying the first half equation by 2 and the second by 5

2MnO4- + 16H+ + 10e --> 2Mn2+ + 8H2O
5NO2- + 5H2O --> 5NO3- + 10e + 10H+
2MnO4- + 16H+ + 5NO2- + 5H2O ---> 2Mn2+ + 8H2O + 5NO3- + 10H+

now cancel out similar species on each side

2MnO4- + 6H+ + 5NO2- ---> 2Mn2+ + 3H2O + 5NO3-

And you will see that there are five nitrite ions for each two manganate ions, so the ratio equivalence is 2.5 : 1.
0
#3
(Original post by charco)
The equivalence comes from the addition of balanced half equations:

manganate ions are oxidising agents:

MnO4- + 8H+ + 5e --> Mn2+ + 4H2O

Nitrite ions are reducing agents:

NO2- + H2O --> NO3- + 2e + 2H+

to add the two half equations together you must make the electrons equal by multiplying the first half equation by 2 and the second by 5

2MnO4- + 16H+ + 10e --> 2Mn2+ + 8H2O
5NO2- + 5H2O --> 5NO3- + 10e + 10H+
2MnO4- + 16H+ + 5NO2- + 5H2O ---> 2Mn2+ + 8H2O + 5NO3- + 10H+

now cancel out similar species on each side

2MnO4- + 6H+ + 5NO2- ---> 2Mn2+ + 3H2O + 5NO3-

And you will see that there are five nitrite ions for each two manganate ions, so the ratio equivalence is 2.5 : 1.
Thank you soo much.. AH I'm a lazy cow.. Why didn't I think of this..!? so would you advise always write an ionic equation for these types of questions?!
0
8 years ago
#4
(Original post by ECullen)
Thank you soo much.. AH I'm a lazy cow.. Why didn't I think of this..!? so would you advise always write an ionic equation for these types of questions?!
yes, unless you actually already know the relationship.
0
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