The process is below ZnCO3s HCl ZnCl2aq CO2g H2Ol ZnCL2a

The process is below:

ZnCO3(s) + HCl --> ZnCl2(aq) + CO2(g) + H2O(l)

ZnCL2(aq) + NaOH(aq) --> Zn(OH)2 + NaCl(aq)

Zn(OH)2(s) + heat --> ZnO(s) + H2O(g)

How many moles of ZnO will be formed if 35.0 g of zinc carbonate is used originally? Thanks!

moles of ZnO=moles of ZnCO3

moles of ZnCO3=35/125

=0.28

b)moles of ZnO formed=18.5/(81)

=0.228

so %yeild=0.228*100/0.28

=81.43 %