to yield 70.5g of NO
2007-03-21 16:19:05 · 2 answers · asked by asusam 1 in Science & Mathematics ➔ Chemistry
to yield 70.5 g of NO...I have the answer and apparently its 134g but I dont know how to get that
2007-03-21 16:51:06 · update #1
If the reaction was 100% yield and 70.5 g of NO was obtained, the moles of NO= 70.5/30 = 2.35 moles. From the reaction statement, a minimum of 2.35 moles of NH3 would be required just to produce this mass of NO. Then 2.35/.298= total moles of NH3 used, or approximately 8 moles. The grams of NH3 are appx. 8x17= 126 grams, of which 17x2.35 are used in the reaction, or appx 39 g. Thus the excess NH3 used is appx 87 g.
2007-03-21 16:30:39 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
4 NH3----> 4 NO so that's balanced. consequently, 0.68g NH3/17 grams in keeping with mole = moles NH3 reacted. 0.96g NO/30 grams in keeping with mole = moles of product The ratio of moles product/moles reactant X one hundred = % yield.
2016-12-15 05:57:54 · answer #2 · answered by cheng 4 · 0⤊ 0⤋