1)
PTHH: \(M_xO_y+2yHNO_3\rightarrow xM\left(NO_3\right)_{\dfrac{2y}{x}}+yH_2O\)
\(n_{M_xO_y}=\dfrac{3,06}{x.M_M+16y}\left(mol\right)\Rightarrow n_{M\left(NO_3\right)_{\dfrac{2y}{x}}}=\dfrac{3,06x}{x.M_M+16y}\left(mol\right)\)
=> \(m_{M\left(NO_3\right)_{\dfrac{2y}{x}}}=\dfrac{3,06x}{x.M_M+16y}\left(M_M+\dfrac{124y}{x}\right)=5,22\left(g\right)\)
=> \(3,06x.M_M+379,44y=5,22x.M_M+83,52y\)
=> \(2,16.x.M_M=295,92y\)
=> \(M_M=\dfrac{137}{2}.\dfrac{2y}{x}\left(g/mol\right)\)
Xét \(\dfrac{2y}{x}=2\Rightarrow M_M=137\left(g/mol\right)\Rightarrow Ba\)
\(\dfrac{x}{y}=\dfrac{1}{1}\) => CTHH: BaO
2)
a) Chất rắn không tan là tạp chất
=> mtạp chất = 0,012 (g)
\(\%m_{tạp.chất}=\dfrac{0,012}{7,05}.100\%=0,17\%\)
b)
\(m_{BaCO_3}=0,209-0,012=0,197\left(g\right)\)
=> \(n_{BaCO_3}=\dfrac{0,197}{197}=0,001\left(mol\right)\)
PTHH: \(BaO+CO_2\rightarrow BaCO_3\)
0,001<-0,001<--0,001
=> \(\%m_{BaO\left(biến.thành.muối.cacbonat\right)}=\dfrac{0,001.153}{7,05-0,012}.100\%=2,17\%\)
mtăng = \(m_{H_2O\left(pư\right)}+m_{CO_2\left(pư\right)}\)
=> \(18.n_{H_2O}+0,001.44=7,184-7,05\)
=> \(n_{H_2O}=0,005\left(mol\right)\)
PTHH: \(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)
0,005<-0,005
\(\%m_{BaO\left(bị.hút.ẩm\right)}=\dfrac{0,005.153}{7,05-0,012}.100\%=10,87\%\)
