Giải thích các bước giải:
nCO2=$\frac{0,6}{24}$=0,025(mol)
nCH3COOH=$\frac{30.40}{100.60}$=0,2(mol)
PTHH: ZnCO3+2CH3COOH→(CH3COO)2Zn+H2O+CO2
0,025 0,05 0,025 0,025
ZnO+2CH3COOH→(CH3COO)2Zn+H2O
0,075 0,15 0,075
=>x=mZnCO3+mZnO
=0,025.125+0,075.81
=9,2(g)
mdd sau pư=9,2+40-0,025.44=48,1(g)
n(CH3COO)2Zn=0,025+0,075=0,1(mol)
=>C% (CH3COO)2Zn=$\frac{0,1.183.100}{48,1}$=38,05%