Question

3H2(g)+N2(g)→2NH3(g)

a. How many moles of NH3 can be produced from 13.5 mol of H2 and excess N2?

b. How many grams of NH3 can be produced from 4.90 mol of N2 and excess H2.

c. How many grams of H2 are needed to produce 12.74 g of NH3?

d. How many molecules (not moles) of NH3 are produced from 1.05×10⁻⁴ g of H2?

Answer 1

342(9)N2g) ay no. of mole of He = 13.6 mo 3 mo Hz procluue 2 mo NH3 from eachon ehen 13.S mo H produtex13.5 9 mol NH Henie, g.o mo NH produreoltrom 13.S mo H2 and excey N b no of mole of N2 .90 mo 4.90 mole N2 precute 2x4.90 9.80 mel NH 9.So mel no. 0f mole of NH podueol Hence nx molan maus maus of NH PD ducel 4.80 mel Y 179/mes 166.6 prooluued fromm .9omol N2 & exceus H, Henie, 166.6 ams of NHs 12-143 mars of NH3 no. of mole of NH 12-74 17g/mas nans 0.75 moo Molon maus 2 mo N procued o.75 mop Ns Proeleueol b x0.7s 1.125 mo H2 by 3mol Hz,hen Henie, maus of H, ntedlcol .125 mol x 2.25 a 24/mad of H are neeoleal to proclue 12.79 NH3. Hence, 2.25 grams

maus of no. of mole of H2 1.05x 10 .05 x10 9 - S S.25 x1omo 2.o /mo S.25Xlo mo NH mol H2 prodlue x.25 xlo 3.5X10S 23 molecude hen I mol 6 022 X1o -S mo 3.S X16 meleule 3.SX16S x 6.022 x lo 18 21.077 x1o moleude 19 = 2.1077 x1o moeule 2.1677X10 moleeSen moleusen NH produrcal no. of mole ules op Hence, proclureal fom .osxlo Decudes of NH3 ane 2.l077 X1o3 Henue, ofHz