a,Ta có:mC2H6O+mC3H8O3=103,5(g)
Mà mC2H6O=$\frac{1}{4}$.mC3H8O3
=>$\left \{ {{mC2H6O=103,5:(1+4)=20,7(g)} \atop {mC3H803=103,5-20,7=82,8(g)}} \right.$
=>$\left \{ {{nC2H6O=\frac{20,7}{46}=0,45(mol)} \atop {nC3H8O3=\frac{82,8}{92}=0,9(mol)}} \right.$
Vậy nC2H6O=0,45mol , nC3H8O3=0,9mol
b,nH2O=$\frac{58,8-34,5}{(2+16)}$ =$\frac{58,8-34,5}{18}$ =1,35(mol)
34,5gX gấp số lần so với 103,5gX là:
34,5:103,5=$\frac{1}{3}$(lần)
=>nC2H6O=0,45.$\frac{1}{3}$=0,15(mol)
nC3H8O3=0,9.$\frac{1}{3}$=0,3(mol0
Vậy nC2H6O=0,15mol , nC3H8O3=0,3mol , nH2O=1,35mol
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