3 CuO + 2 NH3 → 3 Cu + N2 + 3 H2O;
(i). n(CuO)=m(CuO)/(3*Mr(CuO))=90.40/(3*80)=0.376 mol
n(NH3)=m(NH3)/(2*Mr(NH3))=18.10/(2*17)=0.532 mol
n(CuO)<n(NH3), so CuO is the limiting reactant;
(ii). m(N2)=m(CuO)*Mr(N2)/(3*Mr(CuO))=90.40*28/(3*80)=10.546 g
Answer: (i). CuO is the limiting reactant;
(ii). 10.546 grams of N2 will be formed
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