Dd thu được có môi trường trung tính

`=>` Bazơ và axit đều hết

`n_{HCl} = 0,2.0,1 = 0,02(mol)`

`2HCl + Ba(OH)_2 -> BaCl_2 + 2H_2O`

`0,02 -> 0,01`

`-> V = V_{ddBa(OH)_2} = (0,01)/(0,1) = 0,1(l) = 100(ml)`

`Ba(OH)_2 -> 2OH^{-} + Ba^{2+}`

`0,1 -> 0,2`

`-> pH = 14 + log(0,2) ~~ 13,3`

\(Đặt:n_{Na_2CO_3}=a\left(mol\right);n_{K_2CO_3}=b\left(mol\right)\\ Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,3\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}106a+138b=38,2\\a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\\ a.C\%_{ddHCl}=\dfrac{0,6.36,5}{200}.100=10,95\%\\ b.m_{ddB}=38,2+200-0,3.44=225\left(g\right)\\ C\%_{ddKCl}=\dfrac{74,5.2.0,2}{225}.100\approx13,244\%\\ C\%_{ddNaCl}=\dfrac{58,5.2.0,1}{225}.100=5,2\%\)

trả lời giùm mik vs

Bài 1: nH2SO4=0.06 mol; nHNO3=0.2 mol

nCa(OH)2=3a.0,4=1.2a(mol)

nNaOH=0,4.2a= 0.8a (mol)

Quy đổi hỗn hợp về HNO3 => nHNO3=0.2+0.06.2=0.32mol

PTHH: 2HNO3 + Ca(OH)2 ----> Ca(NO3)2 + 2H2O

Mol 2,4a 1,2a

HNO3+ NaOH -----> NaNO3 + H2O

0,8a 0,8a

=> 2,4a + 0,8a=0.32 => a = 0.1 (mol)

\(V_{ddNaOH\left(tổng\right)}=400+200=600\left(ml\right)=0,6\left(l\right)\\ n_{NaOH\left(tổng\right)}=0,4.0,5+0,2.1,5=0,5\left(mol\right)\\ C_{MddNaOH\left(cuối\right)}=\dfrac{0,5}{0,6}\approx0,833\left(M\right)\)

a)

$Na_2O + H_2O \to 2NaOH + H_2$
$2Na + 2H_2O \to 2NaOH + H_2$
b)

$n_{Na} = 2n_{H_2} = 0,05.2 = 0,1(mol)$

$\Rightarrow n_{Na_2O} = \dfrac{14,7 - 0,1.23}{62} = 0,2(mol)$
$n_{NaOH} = n_{Na} + 2n_{Na_2O} = 0,5(mol)$
$C_{M_{NaOH}} = \dfrac{0,5}{0,2} = 2,5M$

c)

$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{H_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,25(mol)$

$m_{dd\ H_2SO_4} = \dfrac{0,25.98}{20\%} = 122,5(gam)$
$V_{dd\ H_2SO_4}  = \dfrac{122,5}{1,4} = 87,5(ml)$

conan cung pai hoi

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