\(n_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\)
\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
0,1_______0,1____________0,1______________
\(m_{muoi}=0,1.\left(12+3+12+16+16\right).2+65=18,3\left(g\right)\)
\(\Rightarrow V_{H2}=0,1.22,4=2,24\left(l\right)\)
Zn+CH3COOH--->H2+(CH3COO)2Zn
n Zn=6,5/65=0,1(mol)
n H2=n Zn=0,1(mol)
V H2=0,1.22,4=2,24(l)
n (CH3COO)2Zn=n Zn=0,1(mol)
m (CH3COO)2Zn=0,1.183=18,3(g)