1)даноn(CH3COOH)=0.6 molNa-----------------------V(H2)-? 0.6 mol Xmol2CH3COOH+2Na-->2CH3COONa+H2↑2mol 1mol0.6/2=X/1X=0.3 molV(H2)=n*Vm=0.3*22.4=6.72 лответ 6.72 л2)даноn(CH3COOH)=0.75 molNaη=0.8-----------------------m(практCH3COONa)-? 0.75 mol Xmol2CH3COOH+2Na-->2CH3COONa+H2↑n(CH3COOH)=n(CH3COONa)=0.75 molM(CH3COONa)=82 g/molm(CH3COONa)n*M=0.75*82=61.5g - теоретический выходm(пракCH3COONa)=m(теор)*η=61.5*0.8=49.2 gответ 49.2 g