nAl=m/M=5,4/27≈0,2 mol
2Al+6CH3COOH→2(CH3COO)3Al+3H2
2 6 2 3
0,2 0,6 0,2 0,3 (mol)
a) mctCH3COOH=n×M=0,6×60=36g
mddCH3COOH=(mct×100%)/C%=(36×100%)/4%=900g
b) m(CH3COO)3Al=n×M=0,2×204=40,8g
c) VH2=n×22,4=0,3×22,4=6,72l
d) mH2=n×M=0,3×2=0,6g
mddspư=mAl+mddCH3COOH-mH2
=5,4+900-0,6
=904,8g
C%ddspư=(mmuối×100%)/mddspư=(40,8×100%)/904,8≈5%