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
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.
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
117,514 g NH3 are obtained.
The answer is 123,64 g ammonia.
115,811 g ammonia are produced.
The mass of ammonia is 121,9 g.
4
The mass of nitrous oxide is 262,8 g.
Nothing is produced, 500g potassium chlorate will be the same 500 g potassium chlorate after reaction. Actually there is no reaction at all.
Following the Law of Conservation of Mass (see link below), there will be 20 grams of products in a reaction of 20 grams of reactions.
Depends on what the last reaction is. And probably what the first reaction was too. Insufficient information in the question.
Two moles of water are produced.
4
26.20
The mass of nitrous oxide is 262,8 g.
Nothing is produced, 500g potassium chlorate will be the same 500 g potassium chlorate after reaction. Actually there is no reaction at all.
Following the Law of Conservation of Mass (see link below), there will be 20 grams of products in a reaction of 20 grams of reactions.
Depends on what the last reaction is. And probably what the first reaction was too. Insufficient information in the question.
Answer= 4 mol H3PO4 - White Plains Public Schools
To calculate the grams of dichloromethane produced, we first need to convert the mass of methane from kilograms to grams, which is 1,540 grams. Given a yield of 48.2%, we can multiply this by the yield percentage to find the actual amount of dichloromethane produced: 1,540 grams of methane x 0.482 = 742.28 grams of dichloromethane.
51.4 - 51.8
c. 16.0 grams CH4/1 mol C04
96.75 grams of NaCl