A mixture of zinc oxide and zinc weight 8,525 g was

given:
m(mixture) = 8,525 g
V(Н2) = 2,24 lit
find:
(ZnО) -?
Solution:
nbsp;ZnО + 2НСl nbsp;= ZnСl2 nbsp;+ nbsp;2Н2О nbsp;nbsp;nbsp;nbsp;nbsp;(1)
Zn + nbsp;2НСl = ZnСl2 nbsp;+ nbsp;Н2 nbsp;nbsp;nbsp;nbsp;nbsp;nbsp;nbsp;nbsp;nbsp;nbsp;nbsp;(2)
(Н)=(V(Н))/V_m = (2,24 lit)/(22,4 l/mol)=0,1 mol
(Н2) = (Zn) nbsp; (Zn) nbsp;nbsp;0,1 mol
М(Zn) = 65 g/mol
m(Zn) = (Zn) М(Zn) = 0,1 mol 65 g/mol nbsp;= 6,5 g
m(ZnО) = m(mixture) - m(Zn) nbsp;= 8,525 g 6,5 g = 2,065 g
(ZnО)=(m(ZnО))/(m(mixture))= (2,025 g)/(8,525 g)=0,24,либо 24 %
Answer:(ZnО) = 24 %.