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幼苗
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设21mL0.5mol/L烧碱溶液中和H2SO4为xmol:
H2SO4 + 2NaOH = Na2SO4 + 2H2O
1 2
x 0.021×0.5
解得:x = 0.00525
与NH3反应的H2SO4为:0.025L×0.25mol/L - 0.00525mol = 0.001mol
设反应产生NH3为ymol:
2NH3 + H2SO4 = (NH4)2SO4
2 1
y 0.001
解得:y = 0.002
0.002molNH3含N原子为0.002mol
样品中含氮元素为:0.002mol × 14g/mol = 0.028g
样品中氮元素质量分数为:0.028g ÷ 0.2102g × 100% = 13.3%
1年前
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