Đáp án:
PTHH: `Zn + 2HCl → ZnCl2 + H2↑`
a) `n_(Zn)=(6,5)/65=0,1 (mol)`
Theo PT: `n_(H2)=n_(Zn)=0,1 (mol)`
`→V_(H2(đktc)``=n_(H2).22,4=0,1.22,4=2,24 (lít)`
b) Theo PT: `n_(HCl)=2.n_(Zn)=2.0,1=0,2 (mol)`
`→ m_(HCl)=0,2.36,5=7,3 (g)`
c) PTHH: `CuO + H2`$\xrightarrow{t^o}$ `Cu +H2O (2)`
Theo PT `(2): n_(Cu)=n_(H2)=0,1 (mol)`
`→ m_(Cu)=0,1.64=6,4 (g)`