Zn+2HCl-->ZnCl2+H2
0,1-0,2------------------0,1 mol
nZn=6,5\65=0,1 mol
=>mHCl=0,2.36,5=7,3 g
=>VH2=0,1.22,4=2,24 l
\(Zn+2HCl-->ZnCl2+H2\)
b) \(n_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\)
\(n_{HCl}=2n_{Zn}=0,2\left(mol\right)\)
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
c) \(n_{H2}=n_{Zn}=0,1\left(mol\right)\)
\(V_{H2}=0,1.22,4=2,24\left(l\right)\)