\(a,n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH: \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
0,2----->0,4----------------->0,2--------------->0,2
\(b,V_{H_2}=0,2.22,4=4,48\left(l\right)\\ c,m_{muối}=0,2.142=28,4\left(g\right)\\ d,C_{M\left(CH_3COOH\right)}=\dfrac{0,4}{0,5}=0,8M\)
