a)
\(n_{M_xO_y}=\dfrac{3,06}{x.M_M+16y}\left(mol\right)\)
PTHH: \(M_xO_y+2yHNO_3\rightarrow xM\left(NO_3\right)_{\dfrac{2y}{x}}+yH_2O\)
\(n_{M\left(NO_3\right)_{\dfrac{2y}{x}}}=\dfrac{5,22}{M_M+\dfrac{124y}{x}}\left(mol\right)\)
Theo PTHH: \(x.n_{M_xO_y}=n_{M\left(NO_3\right)_{\dfrac{2y}{x}}}\)
=> \(\dfrac{3,06x}{x.M_M+16y}=\dfrac{5,22}{M_M+\dfrac{124y}{x}}\)
=> \(3,06.x.M_M+379,44y=5,22.x.M_M+83,52y\)
=> \(M_M=68,5.\dfrac{2y}{x}\left(g/mol\right)\)
Xét \(\dfrac{2y}{x}=2\) thỏa mãn => MM = 137 (g/mol)
=> M là Ba
b)
Chất rắn không tan sau khi hòa tan cặn vào dd HCl là tạp chất
=> \(\%m_{tạp.chất}=\dfrac{0,012}{7,05}.100\%=0,17\%\)
Ta có: \(m_{BaCO_3}=0,209-0,012=0,197\left(g\right)\Rightarrow n_{BaCO_3}=\dfrac{0,197}{197}=0,001\left(mol\right)\)
PTHH: \(BaO+CO_2\rightarrow BaCO_3\)
0,001<--0,001<--0,001
=> \(\%m_{BaO\left(biến.thành.muối.cacbonat\right)}=\dfrac{0,001.153}{7,05-0,012}.100\%=2,174\%\)
\(m_{tăng}=m_{CO_2}+m_{H_2O}\)
=> \(0,001.44+18.n_{H_2O}=7,184-7,05\)
=> \(n_{H_2O}=0,005\left(mol\right)\)
PTHH: \(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)
0,005<--0,005
=> \(\%m_{BaO\left(bị.hút.ẩm\right)}=\dfrac{0,005.153}{7,05-0,012}.100\%=10,87\%\)
