Cho 14,95 gam hỗn hợp A gồm kim

Chọn đáp án B

$A\left\{\begin{array}{l}\underset{x\text{\hspace{0.33em}mol}}{\underbrace{M}}\\ \underset{y\text{\hspace{0.33em}mol}}{\underbrace{M_{2}O}}\\ \underset{\mathrm{0,05}\text{\hspace{0.33em}mol}}{\underbrace{M_{2}C{O}_3}}\end{array}\right.\stackrel{+H_{2}O}{\to }\left\{\begin{array}{l}\underset{(x+2y)}{\underbrace{MOH}}\\ \underset{\mathrm{0,05}}{\underbrace{M_{2}C{O}_3}}\end{array}\right.\stackrel{+HCl:\mathrm{0,45}\text{\hspace{0.33em}mol}}{\to }\left|\begin{array}{l}C{O}_2\stackrel{+Ca{\left(OH\right)}_2:\mathrm{0,035}}{\to }\left\{\begin{array}{l}CaC{O}_3\downarrow :\mathrm{0,02}\\ Ca{\left(HC{O}_3\right)}_2\end{array}\right.\\ MCl\stackrel{BTNT.Cl}{\to }{n}_{MCl}=\mathrm{0,45}\end{array}\right.$

$\begin{array}{l}{n}_{C{O}_2}=n_{O{H}^{-}}-n_{CaC{O}_3}=\mathrm{2.0,035}-\mathrm{0,02}=\mathrm{0,05}\text{\hspace{0.33em}mol; }\stackrel{BTNT.C}{\to }{n}_{M_{2}C{O}_3}=\mathrm{0,05}\text{\hspace{0.33em}mol}\\ \stackrel{BTNT.M}{\to }{n}_M=(x+2y)+\mathrm{2.0,05}=\mathrm{0,45}\to x+2y=\mathrm{0,35}\text{ (1)}\end{array}$

$\begin{array}{l}\to y=\mathrm{0,175}-\mathrm{0,5}x\to 0<y<\mathrm{0,175}\text{ (2) }\\ m_A=m_M+m_{O\left(M_{2}O\right)}+m_{C{O}_3\left(M_{2}C{O}_3\right)}=\mathrm{0,45}M+16y+\mathrm{60.0,05}=\mathrm{14,95}\text{ (3) }\end{array}$

$\to \text{M = }\frac{\mathrm{11,95}-16y}{\mathrm{0,45}}\text{ (4)}$

$\text{Tu (2) va (4)}\to \frac{\mathrm{11,95}-\mathrm{16.0,175}}{\mathrm{0,45}}<M<\frac{\mathrm{11,95}-16.0}{\mathrm{0,45}}$

$\to \mathrm{20,33}<M<\mathrm{26,55}\text{ }\to$Kim loại kiềm M là Na

$Tu\text{ (1) va (3)}\to \left\{\begin{array}{l}x+2y=\mathrm{0,35}\\ \mathrm{0,45.23}+16y+\mathrm{60.0,05}=\mathrm{14,95}\end{array}\right.\to \left\{\begin{array}{l}x=\mathrm{0,15}\\ y=\mathrm{0,1}\end{array}\right.$

$\to \%m_{N{a}_{2}O}=\frac{\mathrm{62.0,1}}{\mathrm{14,95}}.100\%=\mathrm{41,47}\%$