# What mass of NH3(g) will be formed when 50 kg of N2(g) and 10.0 kg of H2(g) are mixed?

1
by ishitab

2014-12-27T14:57:19+05:30

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N₂ +  3  H₂  ===>  2 N H₃

1 mole of N₂ (atomic number 14, so 28 grams) and 3 moles of Hydrogen (atomic number 1, so 2 grams)  will combine to give  2 moles of Ammonia (34 grams).

the proportion of combination is  1 mole of Nitrogen and 3 moles of Hydrogen.

We are given  50 Kg = 50,000/28 = 1785.xx moles of Nitrogen and 10 kg = 5,000 moles of Hydrogen.

The amount of Hydrogen we have is less than 3 * number of moles of Nitrogen present.  So we will take the amount of Hydrogen present and take 1/3 number of moles of Hydrogen for the reaction.  Hence,  some moles of Nitrogen will be left out after the reaction.

Number of moles in reaction:  5,000 moles of Hydrogen + 5,000/3 moles of Nitrogen = 2 * 5, 000 /3 moles of Ammonia.

Mass of Ammonia formed = 2 / 3 * 5, 000 * (14+3*1) = 56.667 kg

Moles of Nitrogen left out = 50, 000 / 28 -  5, 000 / 3 = 119.05 moles
Mass of Nitrogen left out = 3.333 kg

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You can see that total mass of reactants = 50 + 10 = 60 kg
products =  56.667 kg of ammonia and 3.333 kg of  Nitrogen

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