Bài 1:

$a,n_{FeCl_2}=\dfrac{48,75}{162,5}=0,3(mol)\\ PTHH:3NaOH+FeCl_3\to Fe(OH)_3\downarrow+3NaCl\\ \Rightarrow n_{Fe(OH)_3}=n_{FeCl_3}=0,3(mol)\\ \Rightarrow m_{Fe(OH)_3}=0,3.107=32,1(g)\\ b,n_{NaOH}=3n_{FeCl_3}=0,9(mol)\\ \Rightarrow m_{dd_{NaOH}}=\dfrac{0,9.40}{10\%}=360(g)\\ c,n_{NaCl}=n_{NaOH}=0,9(mol)\\ \Rightarrow C\%_{NaCl}=\dfrac{0,9.58,5}{360+48,75-32,1}.100\%=13,98\%\\ d,PTHH:2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{3}{2}n_{Fe(OH)_3}=0,45(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,45.98}{20\%}=220,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{220,5}{1,14}\approx 193,42(ml)$

Câu 2:

$a,n_{Na_2SO_4}=\dfrac{142.10\%}{100\%.142}=0,1(mol)\\ PTHH:Na_2SO_4+Ba(OH)_2\to 2NaOH+BaSO_4\downarrow\\ \Rightarrow n_{BaSO_4}=n_{Na_2SO_4}=0,1(mol)\\ \Rightarrow m_{BaSO_4}=0,1.233=23,3(g)\\ b,n_{Ba(OH)_2}=n_{Na_2SO_4}=0,1(mol)\\ \Rightarrow m_{dd_{Ba(OH)_2}}=\dfrac{0,1.171}{15\%}=114(g)\\ c,n_{NaOH}=2n_{BaSO_4}=0,2(mol)\\ \Rightarrow C\%_{NaOH}=\dfrac{0,2.40}{142+114-23,3}.100\%=3,44\%$

Bài 3:

$500(ml)=0,5(l)\\ a,PTHH:K_2O+H_2O\to 2KOH\\ n_{K_2O}=\dfrac{1,88}{94}=0,02(mol)\\ b,n_{KOH}=2n_{K_2O}=0,04(mol)\\ \Rightarrow C_{M_{KOH}}=\dfrac{0,04}{0,5}=0,08M\\ c,PTHH:KOH+HCl\to KCl+H_2O\\ \Rightarrow n_{KOH}=n_{HCl}=0,02(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,02.36,5}{7,3\%}=10(g)$

Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!

Giải thích các bước giải:

\(\begin{array}{l}
1,\\
3NaOH + FeC{l_3} \to Fe{(OH)_3} + 3NaCl\\
{n_{FeC{l_3}}} = 0,3mol\\
 \to {n_{Fe{{(OH)}_3}}} = {n_{FeC{l_3}}} = 0,3mol\\
 \to {m_{Fe{{(OH)}_3}}} = 32,1g\\
{n_{NaOH}} = 3{n_{FeC{l_3}}} = 0,9mol\\
 \to {m_{NaOH}} = 36g\\
 \to {m_{NaOH}}{\rm{dd}} = \dfrac{{36 \times 100}}{{10}} = 360g\\
{n_{NaCl}} = 3{n_{FeC{l_3}}} = 0,9mol\\
 \to {m_{NaCl}} = 52,65g\\
{m_{{\rm{dd}}}} = 360 + 48,75 - 32,1 = 376,65g\\
 \to C{\% _{NaCl}} = \dfrac{{52,65}}{{376,65}} \times 100\%  = 13,98\% \\
2Fe{(OH)_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 6{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{3}{2}{n_{Fe{{(OH)}_3}}} = 0,45mol\\
 \to {m_{{H_2}S{O_4}}} = 44,1g\\
 \to {m_{{H_2}S{O_4}}}{\rm{dd}} = \dfrac{{44,1 \times 100}}{{20}} = 220,5g\\
 \to {V_{{H_2}S{O_4}}} = \dfrac{{220,5}}{{1,14}} = 193,42ml
\end{array}\)

\(\begin{array}{l}
2,\\
N{a_2}S{O_4} + Ba{(OH)_2} \to BaS{O_4} + 2NaOH\\
{n_{N{a_2}S{O_4}}} = \dfrac{{142 \times 10}}{{100 \times 142}} = 0,1mol\\
 \to {n_{BaS{O_4}}} = {n_{N{a_2}S{O_4}}} = 0,1mol\\
 \to {m_{BaS{O_4}}} = 23,3g\\
{n_{Ba{{(OH)}_2}}} = {n_{N{a_2}S{O_4}}} = 0,1mol\\
 \to {m_{Ba{{(OH)}_2}}} = 17,1g\\
 \to {m_{Ba{{(OH)}_2}}}dd = \dfrac{{17,1 \times 100}}{{15}} = 114g\\
{n_{NaOH}} = 2{n_{N{a_2}S{O_4}}} = 0,2mol\\
{m_{dd}} = 142 + 114 - 23,3 = 232,7g\\
 \to C{\% _{NaOH}} = \dfrac{{0,2 \times 40}}{{232,7}} \times 100\%  = 3,44\% 
\end{array}\)

\(\begin{array}{l}
3,\\
{K_2}O + {H_2}O \to 2KOH\\
{n_{{K_2}O}} = 0,02mol\\
 \to {n_{KOH}} = 2{n_{{K_2}O}} = 0,04mol\\
 \to C{M_{KOH}} = \dfrac{{0,04}}{{0,5}} = 0,08M\\
KOH + HCl \to KCl + {H_2}O\\
{n_{KOH}} = 0,02mol\\
 \to {n_{HCl}} = {n_{KOH}} = 0,02mol\\
 \to {m_{HCl}} = 0,73g\\
 \to {m_{HCl}}dd = \dfrac{{0,73 \times 100}}{{7,3}} = 10g
\end{array}\)