A 0.568-g sample of fertilizer contained nitrogen as ammonium sulfate, . It was analyzed for nitrogen by heating with sodium hydroxide. The ammonia was collected in 45.2 mL of 0.192 M (hydrochloric acid), with which it reacted. This solution was titrated for excess hydrochloric acid with 42.4 mL of 0.133 M . What is the percentage of nitrogen in the fertilizer

7.49% of nitrogen

Explanation:

The reaction of ammonium sulfate ((NH₄)₂SO₄) with NaOH is:

(NH₄)₂SO₄ + NaOH → Na₂SO₄ + 2 H₂O + 2 NH₃

Then, the ammonia is collected in HCl, thus:

NH₃ + HCl → NH₄Cl

And the excess of HCl is neutralized with NaOH, thus:

HCl + NaOH → H₂O + NaCl

That means moles of NH₃ are the difference between initial moles of HCl and the excess moles of HCl.

Initial moles of HCl are:

0.0452L × (0.192mol / L) = 8.68x10⁻³ moles of HCl.

Excess moles are:

0.0424L × (0.133mol / L) = 5.64x10⁻³ moles of HCl.

The difference is:

8.68x10⁻³ moles of HCl - 5.64x10⁻³ moles of HCl =

3.04x10⁻³ moles of HCl ≡ moles of NH₃ ≡ moles of nitrogen.

As nitrogen weights 14.0g/mol:

3.04x10⁻³ moles N × (14.0g / mol) = 0.04256g of nitrogen

Percentage of nitrogen is:

(0.04256g of nitrogen / 0.568g) × 100 = 7.49% of nitrogen

catya that's not even a b or c

the correct answer to this question is 2