\(n_{ZnO}=\dfrac{8,1}{81}=0,1mol\)
\(ZnO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2O\)
0,1 0,2 0,1 ( mol )
\(m_{\left(CH_3COO\right)_2Zn}=0,1.183=18,3g\)
\(m_{CH_3COOH}=0,2.60=12g\)
\(C\%_{CH_3COOH}=\dfrac{12}{200}.100=6\%\)
Zn+2CH3COOH->(CH3COO)2Zn+H2
0,1----------------0,2-------0,1
n ZnO=0,1 mol
=>m (CH3COO)2Zn=0,1.183=18,3g
=>C% =\(\dfrac{0,2.60}{200}100\)=6%