A commerical brand of household ammonia was analysed to determine its ammonia content. 18.25g of the household ammonia was dissolved and diluted to 250ml. A 25ml portion of this solution required 23.8ml of 0.105molL-1 HCl for neutralisation.
Calculate the percentage of ammonia in the commercial household ammonia.
I got 4.8%, is this right?
2007-04-04 15:11:51 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
NOT BEING SEXUAL EVER, BE A BEAVER CLEVER AND SQUARE.chemistry question
2007-04-04 15:15:30 · answer #1 · answered by fddd s 2 · 0⤊ 0⤋
0.105 M * 0.0238 L = 0.0025 moles HCl, so there are 0.0025 moles of NH3 in the 25 ml aliquot of ammonia solution. So there would be 0.025 moles in the 250 ml.
0.025 moles * 17 g/mole = 0.425 g NH3
0.425/18.25 * 100 = 2.33%
2007-04-04 22:19:58 · answer #2 · answered by TheOnlyBeldin 7 · 0⤊ 0⤋