a) Ta có:
\(M_{C_2H_5OH}=12.2+1.5+16+1=46\left(\frac{g}{mol}\right)\)
\(\%m_O=\frac{16}{46}.100=34,783\%\)
Trong 9,2 g C2H5OH, O chiếm:
\(m_O=m_{C_2H_5OH}.\%m_O=9,2.34,783\%=3,2\left(g\right)\)
b) Ta có:
\(M_{SO_3}=32+3.16=80\left(\frac{g}{mol}\right)\)
\(\%m_O=\frac{3.16}{80}.100=60\%\)
Khối lượng O có trong 16g SO3:
\(m_O=\%m_O.m_{SO_3}=60\%.16=9,6\left(g\right)\)
c) Ta có:
\(M_{Al_2\left(SO_4\right)_3}=2.27+32.3+3.4.16=342\left(\frac{g}{mol}\right)\)
=> \(\%_O=\frac{4.3.16}{342}.100\approx56,14\%\)
Trong 17,1 g Al2(SO4)3 , O chiếm:
\(m_O=\%m_O.m_{Al_2\left(SO_4\right)_3}=56,14\%.17,1=9,59994\left(g\right)4\)