\(a,n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PTHH: Mg + 2CH3COOH ---> (CH3COO)2Mg + H2
0,1---->0,2-------------------->0,1------------>0,1
\(\rightarrow m_{ddCH_3COOH}=\dfrac{0,2.60}{20\%}=60\left(g\right)\)
b, mH2 = 0,1.2 = 0,2 (g)
c, mdd sau phản ứng = 60 + 2,4 - 0,2 = 62,2 (g)
\(\rightarrow C\%_{\left(CH_3COO\right)_2Mg}=\dfrac{0,1.142}{62,2}.100\%=22,83\%\)
\(n_{Zn}=\dfrac{2,4}{24}=0,1mol\)
\(Zn+CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+\dfrac{1}{2}H_2\)
0,1 0,1 0,1 0,05 ( mol )
\(m_{dd_{CH_3COOH}}=\dfrac{0,1.60}{20\%}=30g\)
\(V_{H_2}=0,05.22,4=1,12l\)
\(C\%_{\left(CH_3COO\right)_2Zn}=\dfrac{0,1.183}{2,4+30-0,05.2}.100\%=56,65\%\)