A mixture of zinc oxide and zinc weight 8,525 g was treated with excess hydrochloric acid, wherein stood 2,24 liters of gas (normal conditions

given:
m(mixture) = 8,525 g
V(Н2) = 2,24 lit
find:
ω(ZnО) -?
Solution:
 ZnО + 2НСl  = ZnСl2  +  2Н2О      (1)
Zn +  2НСl = ZnСl2  +  Н2↑            (2)
ν(Н₂)=(V(Н₂))/V_m = (2,24 lit)/(22,4 l/mol)=0,1 mol
ν(Н2) = ν(Zn)  ⇒ ν(Zn)   0,1 mol
М(Zn) = 65 g/mol
m(Zn) = ν(Zn) • М(Zn) = 0,1 mol • 65 g/mol  = 6,5 g
m(ZnО) = m(mixture) - m(Zn)  = 8,525 g – 6,5 g = 2,065 g
ω(ZnО)=(m(ZnО))/(m(mixture))= (2,025 g)/(8,525 g)=0,24,или 24 %
Answer:ω(ZnО) = 24 %.