3H_2(g)+N_2(g) -----> 2NH_3(g) ?
Hydrogen gas, H_2, reacts with nitrogen gas,N_2, to form ammonia gas, NH_3, according to the equation
3H_2(g)+N_2(g) -----> 2NH_3(g)
1.How many grams of H_2 are needed to produce 11.69 g of NH_3?
2.How many molecules (not moles) of NH_3 are produced from 8.65×10^-4 g of H_2?
With explain plz!
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Verified answer
1.
moles NH3 = 11.69 g/ 17.0307 g/mol= 0.686
the ratio between H2 and NH3 is 3 : 2
moles H2 needed = 0.686 x 3 / 2= 1.03
mass H2 = 1.03 mol x 2.016 g/mol= 2.08 g
2.
Moles H2 = 8.65 x 10^-4 g/ 2.016 g/mol= 0.000429
moles NH3 = 0.000429 x 2 / 3 =0.000286
molecules = 0.000286 x 6.02 x 10^23 = 1.72 x 10^20