Đáp án:

\(\begin{array}{l}
1.\\
a)C{\% _{NaOH}} = 10\% \\
b)C{\% _{N{a_2}S{O_4}}} = 10,92\% \\
2.\\
a){m_{{\rm{dd}}HCl}} = 100g\\
b){V_{C{O_2}}} = 4,48l\\
c)C{\% _{NaCl}} = 8,04\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1.\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
a)\\
{m_{{H_2}S{O_4}}} = \dfrac{{200 \times 19,6\% }}{{100\% }} = 39,2g\\
 \to {n_{{H_2}S{O_4}}} = 0,4mol\\
 \to {n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,8mol\\
 \to {m_{NaOH}} = 32g\\
 \to C{\% _{NaOH}} = \dfrac{{32}}{{320}} \times 100\%  = 10\% \\
b)\\
{n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,4mol\\
 \to {m_{N{a_2}S{O_4}}} = 56,8g\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}{H_2}S{O_4}}} + {m_{{\rm{dd}}NaOH}} = 520g\\
 \to C{\% _{N{a_2}S{O_4}}} = \dfrac{{56,8}}{{520}} \times 100\%  = 10,92\% \\
2.\\
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
a)\\
{m_{N{a_2}C{O_3}}} = \dfrac{{200 \times 10,6\% }}{{100\% }} = 21,2g\\
 \to {n_{N{a_2}C{O_3}}} = 0,2mol\\
 \to {n_{HCl}} = 2{n_{N{a_2}C{O_3}}} = 0,4mol\\
 \to {m_{HCl}} = 14,6g\\
 \to {m_{{\rm{dd}}HCl}} = \dfrac{{14,6}}{{14,6\% }} \times 100\%  = 100g\\
b)\\
{n_{C{O_2}}} = {n_{N{a_2}C{O_3}}} = 0,2mol\\
 \to {V_{C{O_2}}} = 4,48l\\
c)\\
{n_{NaCl}} = 2{n_{N{a_2}C{O_3}}} = 0,4mol\\
 \to {m_{NaCl}} = 23,4g\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}N{a_2}C{O_3}}} + {m_{{\rm{dd}}HCl}} - {m_{C{O_2}}} = 291,2g\\
 \to C{\% _{NaCl}} = \dfrac{{23,4}}{{291,2}} \times 100\%  = 8,04\% 
\end{array}\)

Đáp án:

 Bài 1 

a) PTHH : `H_2SO_4 + 2NaOH → Na_2SO_4 + 2H_2O`

`mctH_2SO_4=frac{19,6.200}{100}=39,2`g

`nH_2SO_4=frac{39,2}{98}=0,4` mol

`⇒nNaOH=2nH_2SO_4=0,8`mol

`⇒mNaOH=0,8.40=32`g

`⇒C%NaOH=frac{32}{320}.100=10%`

b) dd A tức là dd `Na_2SO_4`

ta có : `nNa_2SO_4=nH_2SO_4=0,4`mol

`⇒mNa_2SO_4=142.0,4=56,8`g

`⇒mdd=520`g

`C%Na_2SO_4=frac{56,8}{520}.100≈10,92%`

Bài 2 :

a) PTHH : `Na_2CO_3 + 2HCl → 2NaCl + H_2O + CO_2↑`

`mctNa_2CO_3=frac{10,6.200}{100}=21,2` g

`nNa_2CO_3=frac{21,2}{106}=0,2`mol

`⇒nHCl=2nNa_2CO_3=0,4` mol

`⇒mHCl=0,4.36,5=14,6`g

`mddHCl=frac{14,6.100}{14,6}=100`g

b)`nCO_2=0,2`mol

`⇒VCO_2=0,2.22,4=4,48` lít

c)`nNaCl=2nNa_2CO_3=0,4`mol

`mNaCl=0,4.58,5=23,4`g

`mdd=291,2`g

`⇒C%NaCl=frac{23,4}{291,2}.100≈8,04%`

$\text{*Khiên}$

Giải thích các bước giải: