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34 grams of Ammonia
45 g water are obtained.
The mass of ammonia is 147,5 g.
76 g ammonia are obtained.
The gram molecular mass of ammonia is 17.03. The formula shows that only one atom of nitrogen is required for each mole of ammonia; 4.12 mol of diatomic nitrogen contains 8.24 mol of nitrogen atoms, and with excess H, all of this nitrogen can be converted to ammonia. Therefore, 8.24 mol of ammonia can be produced, and multiplying this number by17.03 yields a total mass of 140.3 grams of ammonia, to the justified number of significant digits.
34 grams of Ammonia
Balanced equation first. N2 + 3H2 >> 2NH3 (hydrogen is limiting and drives the reaction ) 3.41 grams H2 (1mol/2.016g )(2mol NH3/3mol H2 )(17.034g NH3/1mol NH3 ) = 19.2 grams of ammonia produced ( this is called the Born-Haber process )
11 grams because all is reacted and there is no reactant left over, although if there were only 3 grams of carbon there would have to be 6 grams of oxygen for this to be viable as carbon dioxide is CO2 so the question asked was itself wrong.
16.0 g
45 g water are obtained.
The mass of ammonia is 147,5 g.
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76 g ammonia are obtained.
89,6 g ammonia are obtained.
How many grams of oxygen are needed to react completely with 200.0 g of ammonia, NH3?4NH3(g) + 5O2(g) => 4NO(g) + 6H2O(g)
888
6.08