M=11.20 g
m(H₂)=0.6854 g
M(H₂)=2.016 g/mol
M(Mg)=24.305 g/mol
M(Zn)=65.39 g/mol
w-?
m(Mg)=wm
m(Zn)=(1-w)m
Zn + 2HCl = ZnCl₂ + H₂
m₁(H₂)=M(H₂)m(Zn)/M(Zn)=M(H₂)(1-w)m/M(Zn)
Mg + 2HCl = MgCl₂ + H₂
m₂(H₂)=M(H₂)m(Mg)/M(Mg)=M(H₂)wm/M(Mg)
m(H₂)=m₁(H₂)+m₂(H₂)
m(H₂)=M(H₂)(1-w)m/M(Zn)+M(H₂)wm/M(Mg)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
m(H₂)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
(1-w)/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
1/M(Zn)-w/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
w(1/M(Mg)-1/M(Zn))=m(H₂)/{M(H₂)m}-1/M(Zn)
w=[m(H₂)/{M(H₂)m}-1/M(Zn)]/(1/M(Mg)-1/M(Zn))
w=0.583 (58.3%)
I don’t understand the question sorry
Answer:

Explanation:
Hello!
In this case, according to the chemical reaction:

We can evidence the 2:1 mole ratio between hydrogen and tin, thus, we perform the following stoichiometric setup to obtain the mass of produced tin:

Best regards!
a) The change here is that metallic iron is converted into ions and copper is deposited. This is called a displacement reaction.
b)
is oxidised in this reaction.
c)$ \mathrm{Fe_{(s)}+ CuSO_{4(aq)} \rightarrow FeSO_{4(aq)} + Cu_{(s)}}$