N2 + 3H2 ??2NH3 complete the problems below if the reaction above has starting of 28.0 g of N2 and 25.0 g of H2?
(a) which is the limiting reactant?
(b) how many grams of ammonia can be produced from these starting amounts?
(C) in the laboratory exercise a students products measures 28.0 g of ammonia- what is the % yield of the lab exercise?
(d) how many grams of the excess reactant are left over?
please step by step thank you
N2+ 3H2 ---> 2NH3
28 g of N2
25 g of H2
a) Limiting reactant is N2, since there are 1 mol of N2 in 28 g of N2 and 12.5 mol of H2 in 25 g of H2, therefore, we will need 1 mol of N2 and 3 mol of H2, there is excess of H2
If we use 1 mol of N2 we will produce 2 mol of NH3, that is 17 g/gmol or 34 grams of NH3
c) if 28 grams were produced:
% yield = real / theoretical * 100% = 28g / 34 g * 100 = 82.35%
Since we use 3 mol of H2, we get 12.5 mol - 3 = 9.5 mol of H2
9.5 mol * 2 g/gmol = 19 grams of H2 are left
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