When Molybdenum(VI) Oxide and Zinc are heated together they react to form
molybdenum (III) Oxide and Zinc (II) Oxide. What mass of Molybdenum (III)
Oxide and Zinc (II) Oxide is formed when 20.0 grams of Molybdenum(VI) Oxide is
reacted with 10.0 grams of Zn?
2MoO2 + Zn = Mo2O3 + ZnO
n(MoO2)=m(MoO2)/M(MoO2)=20/128=0.156 mol
n(Zn)=m(Zn)/M(Zn)=10/65=0.154 mol
n(MoO2) : n(Zn) = 2 : 1, so n(Zn) excess
n(MoO2) : n(Mo2O3) = 2 : 1
n(Mo2O3) = n(MoO2)/2 = 0.156/2 = 0.078 mol
n(MoO2) : n(ZnO) = 2 : 1
n(ZnO) = n(MoO2)/2 = 0.156/2 = 0.078 mol
m(Mo2O3) = 0.078×240 = 18.7 g
m(ZnO) = 0.078×81 = 6.3 g
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