Help please, Moles question :) Watch

ECullen
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#1
Report Thread starter 8 years ago
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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.


Answer:

25 cm3 of 0.02 MnO4-
= 25 cm3 × 10–3 × 2 × 10–2
= 5 × 10–4 mol
1 mol MnO4 = 2.5 mol NO2
5 × 10–4 mol MnO4 = 2.5 × 5 × 10–4 mol NO2
= 1.25 × 10–3 mol
0.00125 mol NO2 are contained in 10.5 cm3 of solution
0.00125//10.5 × 50 mol NO2 are contained in 50 cm3 of solution
= 0.00595 mol
NaNO2 = 23 + 14 + 32 = 69
0.00595 mol = 0.00595 × 69 = 0.411 g
% = 0.411/1000 × 100 = 0.0411%
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charco
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(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.


Answer:

25 cm3 of 0.02 MnO4-
= 25 cm3 × 10–3 × 2 × 10–2
= 5 × 10–4 mol
1 mol MnO4 = 2.5 mol NO2
5 × 10–4 mol MnO4 = 2.5 × 5 × 10–4 mol NO2
= 1.25 × 10–3 mol
0.00125 mol NO2 are contained in 10.5 cm3 of solution
0.00125//10.5 × 50 mol NO2 are contained in 50 cm3 of solution
= 0.00595 mol
NaNO2 = 23 + 14 + 32 = 69
0.00595 mol = 0.00595 × 69 = 0.411 g
% = 0.411/1000 × 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+
------------------------------------------------add-----
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.
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ECullen
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Report Thread starter 8 years ago
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(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+
------------------------------------------------add-----
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?!
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charco
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(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. :yep:
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