Question

If the following process produces a yield of 77.6% sulfur dioxide by mass, how many moles of sulfur dioxide are obtained upon heating 92.87 grams of zinc bisulfite? Zn(HSO3)2(s) →ZnO(s)+ 2SO2(g)+ H2O(g)

Answer 1

Zn(HSO3)2(s) →ZnO(s)+ 2SO2(g)+ H2O(g)

1 mole of Zn(HSO3)2(s) on heating to gives 2 moles of So2

227.5g of Zn(HSO3)2(s) on heating to gives 2*64g of So2

92.87g of Zn(HSO3)2(s) on heating to gives = 2*64*92.87/227.5 = 52.25g of SO2

percentage yiled = Actual yield*100/theoretical yield

77.7                    = Actual yiled*100/52.25

Actual yield               = 77.7*52.25/100   = 40.6g >>>> answer

no of moles of SO2 = W/G.M.Wt   = 40.6/64   = 0.634 moles