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# help would be appreciated for this question watch

1. i have this question to do for hw except i dunno where to begin, so can someone please help.

ammonium salts are widely used as fertilisers. one standard method for tha analysis of ammonium salts is to react them in solution of methanal, HCHO. this forms a neutral organic compund together with an acid which can be titrated with standard alkali. for ammonium nitrate the equation for this reaction is

4 NH4NO3 + 6 HCHO ----> (CH2)6N4 + 4 HNO3 + 6 H20

15g of a fertilser contaning ammonium nitrate as the only ammonium salt was dissolved in water and a solution made up to 1.00 dm3 with pure water.

25cm3 portions of this solution were treated with saturated aqeous methanal solution and allowed to stand for a few minutes.

the liberated nitric acid was then titrated with 0.1dm3 NaOH solution. the volume of NaOH solution used was 22.3cm3.

what percentage by mass of the fertiliser was ammoium nitrate.
2. Find how many moles of NaOH were used to neutralise the nitric acid (you're given the concentration and volume).

Using HNO3 + NaOH ----> H2O + NaNO3 determine how many moles of nitric acid were formed in the reaction.

Using the equation in your first post then relate this number of moles of HNO3 to the moles of ammonium nitrate.

Convert the moles of NH4NO3 to mass using it's RMM - this will then tell you how many grams of ammonium nitrate contained in the 25cm3 solution.

Hopefully from there you can finish it off
3. sorry but i still dont get this more help needed
4. May well have made a mistake - so helpful if someone can verify.

You have 25.0cm^3 of nitric acid, and titre of 22.30cm^3 of 0.1moldm^3 NaOH. So the moles of NaOH is 0.0223*0.1=0.00223mol.

1:1 molar ratio from EvS equation, so you have 0.00223mol of nitric acid in 25.0cm^3. You can assume the same number of moles will be in every 25.0^3 division of the 1.00dm^3 solution. 25*40=1000. So it follows that 0.00223mol*40=0.0892mol of nitric acid in the 1.0dm^3 solution.

The ratio of amonium nitrate to nitric acid is 1:1. So there are also 0.892moles of ammonium nitrate. Translate this into mass: 0.0892*80=7.136g

7.136g/15g=0.4757
0.4757*100=48% (2 sig.figs)

Sorry if i made a mistake.
5. (Original post by LearningMath)
May well have made a mistake - so helpful if someone can verify.

You have 25.0cm^3 of nitric acid, and titre of 22.30cm^3 of 0.1moldm^3 NaOH. So the moles of NaOH is 0.0223*0.1=0.00223mol.

1:1 molar ratio from EvS equation, so you have 0.00223mol of nitric acid in 25.0cm^3. You can assume the same number of moles will be in every 25.0^3 division of the 1.00dm^3 solution. 25*40=1000. So it follows that 0.00223mol*40=0.0892mol of nitric acid in the 1.0dm^3 solution.

The ratio of amonium nitrate to nitric acid is 1:1. So there are also 0.892moles of ammonium nitrate. Translate this into mass: 0.0892*80=7.136g

7.136g/15g=0.4757
0.4757*100=48% (2 sig.figs)

Sorry if i made a mistake.
looks good to me
6. wouldnt you times the 0.00223 by 63 since 63 is the ram of nitric acid instead of 40
7. (Original post by themissingpiece of urhart)
wouldnt you times the 0.00223 by 63 since 63 is the ram of nitric acid instead of 40
That's not the RAM LearningMath has multipled by. You don't need the RAM - because every mole of HNO3 reacts with one moles of NaOH, if you can work out the number of moles NaOH in the 22.3cm3 that'll be the same as the number of moles of HNO3 in the 25cm3 sample.

You want to know the number of moles in the full 1000cm3, which has been divided into 25cm3 portions. 1000/25 = 40, so multiplying the number of moles in your 25cme sample gets you the total number of moles.

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Updated: January 18, 2010
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