Bài 6 :
$S_{Na_2SO_4} = \dfrac{7,2}{80}.100 = 9(gam)$
$C\%_{Na_2SO_4} = \dfrac{S}{S + 100}.100\% = 8,26\%$
Bài 11 :
$2R + 2H_2O \to 2ROH + H_2$
$R_2O + H_2O \to 2ROH$
$2ROH + H_2SO_4 \to R_2SO_4 + 2H_2O$
$n_{H_2SO_4} = 0,2.0,15 = 0,03(mol)$
Theo PTHH : $n_{ROH} = 2n_{H_2SO_4} = 0,06(mol)$
Gọi $n_R = a(mol) ; n_{R_2O} = b(mol)$
Suy ra: $a + 2b = 0,06$
$\Rightarrow $a = 0,06 - 2b$
$m_A = a.R + b(2R + 16) = aR + 2bR + 16b = R(0,06 - 2b) + 2bR + 16b$
$= 0,06R + 16b = 1,62$
$\Rightarrow R = \dfrac{1,62 - 16b}{0,06}$
mà $0 < b < 0,03 \Rightarrow 19 < R < 27$
Suy ra : $R = 23$ thì thoả mãn. Vậy A gồm $Na(a\ mol)$ và $Na_2O(b\ mol)$
Ta có : $23a + 62b = 1,62$ và $a + 2b = 0,06$
Suy ra : a = 0,03 ; b = 0,015$
$m_{Na} = 0,03.23 = 0,69(gam) ; m_{Na_2O} = 0,015.62 = 0,93(gam)$
`S = (7,2)/80 . 100 = 9 ( g )`
`mdd = 7,2 + 80 = 87,2 ( g )`
`C % = (7,2)/(87,2) . 100 = 8,26%`
Bài 6.
\(S_{Na_2SO_4;10^oC}=\dfrac{7,2}{80}.100=9\left(g\right)\)
Bài 11.
Đặt \(\left\{{}\begin{matrix}n_A=x\\n_{A_2O}=y\end{matrix}\right.\) ( mol )
\(2A+2H_2O\rightarrow2AOH+H_2\)
x x ( mol )
\(A_2O+H_2O\rightarrow2AOH\)
y 2y ( mol )
\(n_{H_2SO_4}=0,2.0,15=0,03\left(mol\right)\)
\(2AOH+H_2SO_4\rightarrow A_2SO_4+2H_2O\)
0,06 0,03 ( mol )
\(\Rightarrow n_{AOH}=x+2y=0,06\) ( mol )
\(\Leftrightarrow x=0,06-2y\)
\(m_{hh}=Ax+y\left(2A+16\right)=1,62\)
\(\Leftrightarrow Ax+2Ay+16y=1,62\)
\(\Leftrightarrow A\left(0,06-2y\right)+2Ay+16y=1,62\)
\(\Leftrightarrow0,06A-2Ay+2Ay+16y=1,62\)
\(\Leftrightarrow y=\dfrac{1,62-0,06A}{16}\)
Vì y là số mol nên \(0< y< 0,03\)
`@`\(y>0\) \(\Rightarrow\dfrac{1,62-0,06A}{16}>0\) \(\Leftrightarrow A< 27\) (1)
`@`\(y< 0,03\) \(\Rightarrow\dfrac{1,62-0,06A}{16}< 0,03\) \(\Leftrightarrow A>19\) (2)
(1);(2) `=>` A là Natri ( Na )
