"2NH_3+3CuO\\to N_2+3Cu+3H_2O"
given 3 moles of "NH_3" and 6 moles of CuO
"NH_3: 3mol\/2(coefficient)=1.5"
CuO: 6mol"\/3=2"
limiting reactant is "NH_3" and excess reactant is CuO.
moles of CuO required =(1.5 moles of "NH_3 \\times" 3 moles CuO)"\/" (2 moles "NH_3" )=2.25moles
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