Answer:
6.69%
Explanation:
Given that:
Mass of the fertilizer = 0.568 g
The mass of HCl used in titration (45.2 mL of 0.192 M)
= [tex]0.192*\frac{45.2}{1000}* \frac{36.5}{1 \ mole}[/tex]
= 0.313 g HCl
The amount of NaOH used for the titration of excess HCl (42.4 mL of 0.133M)
= [tex]\frac{44.3 \ mL * 1.0 \ L}{1000 \ mL} *0.133 \ mole/L[/tex]
= 0.0058919 mole of NaOH
From the neutralization reaction; number of moles of HCl is equal with the number of moles of NaOH is needed to complete the neutralization process
Thus; amount of HCl neutralized by 0.0058919 mole of NaOH = 0.0058919 × 36.5 g
= 0.2151 g HCl
From above ; the total amount of HCl used = 0.313 g
The total amount that is used for complete neutralization = 0.2151 g
∴ The amount of HCl reacted with NH₃ = (0.313 - 0.2151)g
= 0.0979 g
We all know that 1 mole of NH₃ = 17.0 g , which requires 1 mole of HCl = 36.5 g
Now; the amount of HCl neutralized by 0.0979 HCl = [tex]\frac{17}{36.5}*0.0979[/tex]
= 0.0456 g
Therefore, the mass of nitrogen present in the fertilizer is:
= [tex]0.0456 \ g \ NH_3 * \frac{1 \ mol \ NH_3 }{17.0 \ mol \ of \ NH_3} * \frac{1 \mol \ (NH_4)_2SO_4}{2 \ mol \ NH_3 } * \frac{2 \ mol \ N }{1 \mol \ (NH_4)_2SO_4}* \frac{14.0 g }{1 \ mol \ N}[/tex]
= 0.038 g
∴ Mass percentage of Nitrogen in the fertilizer = [tex]\frac{0.038 \ g}{0.568 \ g} * 100[/tex]%
= 6.69%
Answer:
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
