\(n_M=\frac{6,5}{M}\left(mol\right)\); \(n_{MO}=\frac{8,1}{M+16}\left(mol\right)\)
PTHH: 2M + O2 --> 2MO
\(\frac{6,5}{M}\) -> \(\frac{3,25}{M}\) (mol) (1)
Mặt khác: 2M + O2 --> 2MO
\(\frac{4,05}{M+16}\) <- \(\frac{8,1}{M+16}\) (mol) (2)
(1)(2) => \(\frac{3,25}{M}\) = \(\frac{4,05}{M+16}\)
=> M = 65 (g/mol)
=> M là Zn ( kẽm)
Câu 2 :
PTHH : \(2M+O_2\rightarrow2MO\)
\(n_M=\frac{m_M}{M_M}=\frac{6,5}{M_M}\left(mol\right)\)
\(n_{MO}=\frac{m_{MO}}{M_{MO}}=\frac{8,1}{M_{MO}}=\frac{8,1}{M_M+16}\left(mol\right)\)
- Theo PTHH : \(n_M=n_{MO}\)
=> \(\frac{6,5}{M_M}=\frac{8,1}{M_M+16}\)
=> \(8,1M_M=6,5\left(M_M+16\right)\)
=> \(8,1M_M=6,5M_M+104\)
=> \(8,1M_M-6,5M_M=104\)
=> \(1,6M_M=104\)
=> \(M_M=65\) ( đvc )
Vậy M là kim loại kẽm .
Câu 1
%O ( CO2) = \(\frac{16.2}{44}.100\%=72,73\%\)
%O ( SO3) = \(\frac{16.3}{80}.100\text{% =60%}\)
%O ( P2O5) = \(\frac{16.5}{142}.100\%=56.34\%\)
%O ( Fe3O4) = \(\frac{16.4}{232}.100\%=27,59\%\)
=> CO2 có lượng oxi cao nhất
Câu 1 :
- Chất có hàm lượng oxi cao nhất là CO2 \(\left(\approx72\%\right)\)