Bài 15 :
`-m_S=n_S.M_S=0,5.32=16(g)`
`-m_{Fe}=n_{Fe}.M_{Fe}=0,6.56=33,6(g)`
`-m_{Fe_2O_3}=n_{Fe_2O_3}.M_{Fe_2O_3}=0,8.160=128(g)`
`⇒m_X=m_S+m_{Fe}+m_{Fe_2O_3}`
`=16+33,6+128=177,6(g)`
Bài giải :
`-n_{CH_4}=\frac{m_{CH_4}}{M_{CH_4}}=\frac{32}{16}=2(mol)`
Mà `n_{H}=4.n_{CH_4}`
`⇒n_{H}=4.2=8(mol)`
`-m_{H}=n_{H}.M_{H}=8.1=8(g)`
`⇒%m_{H}=\frac{8}{32}.100%=25%`