a)
mSO2=6,4 gam\(\rightarrow\) nSO2=\(\frac{6,4}{\text{32+16.2}}\)=0,1 mol \(\rightarrow\) V SO2=0,1.22,4=2,24 lít
nN2=\(\frac{7}{14.2}\)=0,25 mol \(\rightarrow\) V N2=0,25.22,4=5,6 lít
m H2O=\(\frac{3,4}{18}\)=\(\frac{17}{90}\) mol\(\rightarrow\)V H2O=\(\frac{17}{\text{90.22,4}}\)=4,23 lít
b) Ta có: m Na2CO3=0,5.(23.2+12+16.3)=53 gam
nCO2=\(\frac{44,8}{22,4}\)=2 mol \(\rightarrow\) m CO2=2.(12+16.2)=88 gam
nO2=1,5.1023/(6,023.1023)=0,25 mol \(\rightarrow\) mO2=0,25.32=8 gam