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# Moles Help watch

1) In each case work out the limiting reagent and moles of ammonia formed (assuming complete reaction).
N2+3H2 -> 2NH3
a) 3.00 moles of N2 + 8.00 moles of H2
b) 3.00 moles of N2 + 10.0 moles of H2
c) 0.100 moles of N2 + 0.200 moles of H2
d) 0.50 moles of N2 + 2.00 moles of H2
e) 2.0 moles of N2 + 10.0 moles of H2
2. (Original post by itskerry17)
1) In each case work out the limiting reagent and moles of ammonia formed (assuming complete reaction).
N2+3H2 -> 2NH3
a) 3.00 moles of N2 + 8.00 moles of H2
b) 3.00 moles of N2 + 10.0 moles of H2
c) 0.100 moles of N2 + 0.200 moles of H2
d) 0.50 moles of N2 + 2.00 moles of H2
e) 2.0 moles of N2 + 10.0 moles of H2
From the equation you can see that if you have exactly 3 times as many moles of hydrogen as nitrogen you will use all the nitrogen and hydrogen.

If you have less than 3 times as much nitrogen then you don't have enough nitrogen for complete reaction, the nitrogen is the limiting reactant (because you would have unreacted hydrogen left over).

If you have more than 3 times as many moles hydrogen as nitrogen then hydrogen is the limiting reactant (as you will have unreacted nitrogen left over).

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