Can someone give me some directions as to how to reach the answer for this problem:

Question

A 0.608g sample of fertilizer contained nitrogen as ammonium sulfate,
(NH4)2 SO4(s)+2NaOH(aq)-->Na2SO4(aq)+ 2H2O(l)+2NH3(g)

The ammonia was collected in 46.3 mL of 0.213 M HCL (hydrochloric acid), with which it reacted.

NH3(g)+HCL(aq)-->NH4Cl(aq)

This solution was titrated for excess hydrochloric acid with 44.3 mL of 0.128 M NaOH

NaOH(aq)+HCL(aq)NaCL(aq)+H2O(l)

What is the percentage of nitrogen in the fertilizer?

The answer is 9.66%

Answer 1

How much HCl was used to collect the NH3? That will be moles HCl = M x L.
How much excess HCl was there. That will be the excess titration with NaOH. Moles NaOH = moles HCl excess = L NaOH x M NaOH.
Subtract the amount HCl in which the NH3 was collected from the total HCl to give the moles NH3 from the sample.
Then convert moles NH3 to moles (NH4)2SO4 then to moles N (not N2) using the coefficients in the balanced equation. From there go to grams N, then percent N from
percent N = [(grams N)/grams sample]*100
Post your work if you get stuck.

Answer 2

Sorry, but your wording is quite confusing. How would you convert from NH3 to (NH4)2SO4 then to moles N (not N2), I don't know if I understand your reasoning...since you already have the sample mass and molar mass of (NH4)2SO4 so therefore you can work out how many moles it has- so how do the two relate?

Answer 3

1. Find mol of NaOH:
NaOH: 44.3mL * 0.128 mol / 1 L -> 5,6704 * 10⁻³ mol NaOH
Since proportion NaOH and HCl is 1:1, 5,67 * 10⁻³ mol HCl left, after previous NH₃ + HCl reaction.

2. Find initial mol of HCl
HCl: 46.3 mL * 0.213 mol / 1 L -> 9,86 * 10⁻³ mol HCl

3. Find reacted HCl mol: 9,86 * 10⁻³ - 5,67 * 10⁻³ = 4,19 * 10⁻³ mol HCl
NH₃:HCl = 1:1, thus 4,19 * 10⁻³ mol NH₃ was in the first reaction

4. Find mass of nitrogen in 4,19 * 10⁻³ mol NH₃ (4,19 * 10⁻³ mol N)
Molar mass N = 14 g/mol
4,19 * 10⁻³ mol N * 14 g N / 1 mol N -> 58.66 * 10⁻³ g N

5. Find mass percentage
58.7mg / 608mg = 9.65%