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

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

thats a little if u want the answer just look it up on google thats what i do haha : )