ta có phương trình:2K+2CH3COOH=>2CH3COOK+H2
a,mK=0.39(g)=>nK=$\frac{0.39}{39}$=0.01(mol)
=>nH2=0.01/2=0.005(mol)=>vH2=0.002*22.4=0.112(lít)
b,nCH3COOH=0.01(mol)=>mCH3COOH=0.01*60=0.6(g)
c,ta có nCH3COOH=0.01 (mol)=>m CH3COOH=0.01*60=0.6(g)
=>m dd axit cần dùng là $\frac{0.6}{20}$*100=3(g)