\(n_{H^+}=0,01.1+0,01.2.2=0,05\left(mol\right)\)

\(\Rightarrow n_{OH^-}=V\left(2+0,5.2\right)=3V=0,05\Rightarrow V=0,0167\left(l\right)\)

n NaOH = \(\frac{40.20\%}{40}\)= 0,2 (mol)

NaOH + HCl   ---------> NaCl + H2O

  0,2 ---> 0,2                                                   (mol)

=> V_HCl = \(\frac{0,2}{0,5}\)= 0,4 (l)

a, PT: \(Fe+S\underrightarrow{t^o}FeS\)

Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,1}{1}\), ta được Fe dư.

Chất rắn A gồm Fe dư và FeS.

Theo PT: \(n_{Fe\left(pư\right)}=n_{FeS}=n_S=0,1\left(mol\right)\)

\(\Rightarrow n_{Fe\left(dư\right)}=0,2\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(FeS+2HCl\rightarrow FeCl_2+H_2S\)

Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{Fe\left(dư\right)}=0,2\left(mol\right)\\n_{H_2S}=n_{FeS}=0,1\left(mol\right)\end{matrix}\right.\)

Ở cùng điều kiện nhiệt độ và áp suất, %V cũng là % số mol.

\(\Rightarrow\left\{{}\begin{matrix}\%V_{H_2}=\dfrac{0,2}{0,2+0,1}.100\%\approx66,67\%\\\%V_{H_2S}\approx33,33\%\end{matrix}\right.\)

b, Ta có: \(\Sigma n_{HCl\left(dadung\right)}=2n_{Fe}+2n_{FeS}=0,6\left(mol\right)\) (1)

PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Ta có: \(n_{NaOH}=0,1.2=0,2\left(mol\right)\)

\(\Rightarrow n_{HCl\left(dư\right)}=n_{NaOH}=0,2\left(mol\right)\) (2)

Từ (1) và (2) \(\Rightarrow\Sigma n_{HCl}=0,8\left(mol\right)\)

\(\Rightarrow C_{M_{ddHCl}}=\dfrac{0,8}{0,5}=1,6\left(M\right)\)

Bạn tham khảo nhé!

a) PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

                   a_____3a_______a______\(\dfrac{3}{2}a\)   (mol)

                \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

                   b_____2b_______b______b       (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}27a+24b=7,8\\\dfrac{3}{2}a+b=\dfrac{8,96}{22,4}=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2\cdot27=5,4\left(g\right)\\m_{Mg}=2,4\left(g\right)\\n_{HCl}=0,8\left(mol\right)=n_{H^+}\end{matrix}\right.\)

b) PT ion: \(H^++OH^-\rightarrow H_2O\)

                 0,8______0,8

Ta có: \(\left[OH^-\right]=C_{M_{NaOH}}+2C_{M_{Ba\left(OH\right)_2}}=2,2\left(M\right)\) \(\Rightarrow V_{OH^-}=\dfrac{0,8}{2,2}\approx0,36\left(l\right)\)

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

Theo PT:  \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,25\left(mol\right)\)

a, \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)

b, \(m_{CaCO_3}=0,25.100=25\left(g\right)\)

c, \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

Theo PT: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{20\%}=91,25\left(g\right)\)