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Question
Nitrogen and hydrogen react to from ammonia .if 1000 g of hydrogen and 2000 g of nitrogen is reacted.
A) will any if the two react ants remain unreacted?if yes which one and what will be its mass?
B) calculate the mass of ammonia that will be formed
A) will any if the two react ants remain unreacted?if yes which one and what will be its mass?
B) calculate the mass of ammonia that will be formed
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Solution
2 N2 + 3 H2 → 2 NH3
2 moles N2 need 3 moles H2 to form 2 moles NH3
One mole of N2 has 28 g
So 2000 g N2 has 1000/28 = 35.71 moles
and 1000 g H2 has 1000/2 = 500 moles
To find out the limiting reagent divide the no of moles of reactants with their stoichiometric coefficient.
For N2: 35.71/2 = 17.855
For H2: 500/3 = 166.66
The lowest number indicates the limiting reagent which in this case is nitrogen.
B) One mole nitrogen forms 2 moles of ammonia so 35.71 moles will form 2*35.71 moles ammonia.
Mass of ammonia formed = (17*2*35.71) = 1214.14 g.
A) Since hydrogen is in excess it will be left unreacted.
Amount of hydrogen required for 34 g = 6 g
Amount of hydrogen required for 1214.14 g = 6/34*1214.14 g = 214.26 g
Amount left = 785.74 g
2 moles N2 need 3 moles H2 to form 2 moles NH3
One mole of N2 has 28 g
So 2000 g N2 has 1000/28 = 35.71 moles
and 1000 g H2 has 1000/2 = 500 moles
To find out the limiting reagent divide the no of moles of reactants with their stoichiometric coefficient.
For N2: 35.71/2 = 17.855
For H2: 500/3 = 166.66
The lowest number indicates the limiting reagent which in this case is nitrogen.
B) One mole nitrogen forms 2 moles of ammonia so 35.71 moles will form 2*35.71 moles ammonia.
Mass of ammonia formed = (17*2*35.71) = 1214.14 g.
A) Since hydrogen is in excess it will be left unreacted.
Amount of hydrogen required for 34 g = 6 g
Amount of hydrogen required for 1214.14 g = 6/34*1214.14 g = 214.26 g
Amount left = 785.74 g
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