NH3+O2=NO+H2O IF 25G NH3 is reacted with 20g O2, which is the limiting reagent? How much NO will be formed in the reaction?
m(NH3) = 25 g;
M(NH3) = 17 g/mol;
n(NH3) = m(NH3)/M(NH3) = 25/17 = 1.471 mol;
m(O2) = 20 g;
M(O2) = 32 g/mol;
n(O2) = m(O2)/M(O2) = 20/32 = 0.625 mol;
4NH3 + 5O2 = 4NO + 6H2O
By the chemical reaction:
n(NH3)' = n(NH3)/4 = 0.368 mol;
n(O2)' = n(O2)/5 = 0.625/5 = 0.125 mol;
So, O2 is the limiting reagent and NH3 is the excess reagent.
n(NO) = 4/5 * n(O2) = 4/5 * 0.625 = 0.5 mol;
M(NO) = 30 g/mol;
m(NO) = n(NO) * M(NO) = 0.5 * 30 = 15.0 g.
Answer: 15.0 g
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