If 250.3 l of hydrogen gas (d = 0.0899 g/l) reacts with an excess of nitrogen gas, what mass of ammonia would be produced?

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28-09-2017
Chemistry

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If 250.3 l of hydrogen gas (d = 0.0899 g/l) reacts with an excess of nitrogen gas, what mass of ammonia would be produced?

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YellowGold YellowGold · Answer 1 · 2017-10-10T11:26:49+02:00

First convert the given volume of hydrogen gas to grams by multiplying the volume with the given density. The mass of hydrogen gas is 22.50197 grams. Then convert the mass of hydrogen to moles by dividing it with its molar mass. Multiply the moles of hydrogen gas with 2/3 because in the formation of ammonia you would need 3 mol of hydrogen to form 2 moles of ammonia. Then multiply the moles of ammonia with the molar mass which is 17.031 grams. The mass of ammonia is 126.48 grams.