n H2 = 6,72/22,4=0,3(mol)
Zn + 2HCl ----> ZnCl2 + H2
0,3_____________0,3____0,3
ZnO + 2HCl ----> ZnCl2 + H2O
0,1______________0,1
m Zn = 0,3.65=19,5(g) => %m = 19,5/27,6.100%=70,65%
m ZnO = 27,6-19,5=8,1(g) => %m = 100% - 70,65 = 29,35%
n ZnO = 8,1/81 = 0,1 (mol)
m ZnCl2 = (0,1 + 0,3).136=54,4(g)
m dd = 27,6 + 200 - 0,3.2=227(g)
\(C\%=\frac{54,4}{227}.100\%=23,965\%\)