How many grams of NH3 can be produced from the reaction of 17.8 moles of H2 and a sufficient supply of N2 N2 plus 3 H2 and rarr 2 NH3?

Answer 1

I assume you are talking about the Haber Process:

3H2 + N2 ----> 2NH3

If 3.71 mol of N2 is produced, there will be 7.42 mol of ammonia produced (as per mol ratios).

Using the formula, n = g/mw ---> g = 7.42 x 17.034

= 126.39228 grams of NH3

Answer 2

Preliminary calculations: NH3 contains one atom of nitrogen and three atoms of hydrogen in each mole. Therefore, the molecular weight of NH3 is 14.007 + 3(1.008) = 17.031. The fraction of nitrogen by weight in this molecule is 14.007/17.031 = 0.82244.

a. 3.09 moles of nitrogen has a mass of 3.09 X 2.00 X 14.007 = 86.56 grams. From this, 86.56/0.82244 = 105 grams of NH3 can be produced.

b. is the same question as a. and therefore has the same answer!

c. Since the molecule contains only two elements, its fraction by weight of hydrogen is 1 - 0.82244 = 0.17756. The amount of hydrogen needed is therefore 0.17756(13.04) = 2.31 grams, to the justified number of significant digits.

Answer 3

N2 + 3H2 --> 2NH3 ... balanced equation
moles H2 used = 17.8
moles of NH3 theoretically produced = 17.8 moles H2 x 2 moles NH3/3 moles H2 = 11.9 moles NH3

Answer 4

117,514 g NH3 are obtained.

Answer 5

The answer is 123,64 g ammonia.

Answer 6

115,811 g ammonia are produced.

Answer 7

The mass of ammonia is 121,9 g.