A 5.37 g sample of household cleaner is dissolved in 25.0 ml of water to form NH4OH. This solution required 37.3 cm3 of 0.360 M sulphuric acid for neutralization. Calculate the percentage mass of ammonia (NH3) in the cleaner? 2NH4OH + H2SO4 → (NH4)2SO4 + 2H2O
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2NH3(aq) + H2SO4(aq) --> (NH4)2SO4 (aq)
Moles H2SO4 = 0.360 x (37.3/1000) = 1.332 x 10 -2 mol H2SO4 in 25cm3
Moles of NH3 = 2 x 1.332 x 10 -2 = 2.664 x 10 -2 mol in 25cm3
In 250 cm3 = 2.664 x 10 - 1 mol NH3
Mass NH3 = 0.2664 x 17 = 4.57g
% mass = (4.57 / 25.37) x 100 = 18.0%
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