a) CTTQ : RxOy( hóa trị của R là \(\dfrac{2y}{x}\))
=>\(\dfrac{xMR}{xMR+16y}.100=40\%\)
=>MR=\(\dfrac{32y}{3x}=\dfrac{16}{3}.\dfrac{2y}{x}\)
xét => MR=32(g/mol)
=>R :S , RxOy:SO3
b) SO3 +H2O --> H2SO4(1)
2NaOH +H2SO4 --> Na2SO4 +2H2O (2)
nNaOH=\(\dfrac{150.8}{100.40}=0,3\left(mol\right)\)
theo (1,2) : nSO3=1/2nNaOH=0,15(mol)
=>mSO3=12(g)