Câu 1:

\(PTHH:2NaOH+H_2S\rightarrow Na_2S+2H_2O\)

_________________0,05______0,05____________

\(NaOH+H_2S\rightarrow NaHS+H_2O\)

_________0,1 ______0,1________

\(\Rightarrow\Sigma n_{H2S}=0,1+0,05=0,15\left(mol\right)\)

\(\Rightarrow V_{H2S}=0,15.22,4=3,36\left(l\right)\)

Câu 2:

Ta có:

\(n_{NaOH}=0,35.0,1=0,035\left(mol\right)\)

\(n_{H2S}=\frac{0,448}{22,4}=0,02\left(mol\right)\)

\(\Rightarrow T=\frac{n_{NaOH}}{n_{H2S}}=\frac{0,035}{0,02}=1,75\)

\(1< T< 2\Rightarrow\) Tạo cả 2 muối

\(2NaOH+H_2S\rightarrow Na_2S+2H_2O\)

Câu 3:

Ta có:

\(\left\{{}\begin{matrix}n_{Zn}=\frac{13}{65}=0,2\left(mol\right)\\n_S=\frac{4}{32}=0,125\left(mol\right)\end{matrix}\right.\)

\(PTHH:Zn+S\rightarrow ZnS\)

Tỉ lệ : \(\frac{0,2}{1}>\frac{0,125}{1}\Rightarrow\) Zn dư

\(n_{ZnS}=0,125\left(mol\right)\)

\(n_{Zn\left(Dư\right)}=0,2-0,125=0,075\left(mol\right)\)

\(\left\{{}\begin{matrix}n_{H2S}=n_{ZnS}=0,125\left(mol\right)\\n_{H2}=n_{Zn}=0,075\left(mol\right)\end{matrix}\right.\)

\(M_{hh}=\frac{0,125.34+0,075.2}{0,2}=22\)

\(D_{hh/H2}=\frac{22}{2}=11\)

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!

B1:

2NaOH+H₂SO₄\(\rightarrow\)Na₂SO₄+2H₂O

n_NaOH=\(\frac{4}{40}=0.1\)mol

=>n_H2SO4=\(\frac{1}{2}\)n_NaOH=0.05 mol

=>C_M=\(\frac{n_{H2SO42}}{V}\)=\(\frac{0.05}{200}\)=2,5.10^-4 (M)

B2:

Mg+\(\frac{1}{2}\)O₂\(\underrightarrow{t^0}\)MgO                (1)

MgO+2HCl\(\rightarrow\)MgCl₂+H₂O (2)

n_Mg(1)=\(\frac{0,36}{24}=0,015mol\)

=>n_MgO(1)=0,015=n_MgO(2)

n_HCl(2)=2n_MgO(2)=0,03mol

=>C_M(HCl)=\(\frac{n_{HCl}}{V}=\frac{0,03}{100}=3.10^{-4}M\)

a,\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:    0,05     0,1        0,05     0,05

PTHH: MgO + 2HCl → MgCl₂ + H₂O

Mol:       0,1      0,2          0,1

⇒ m_Mg = 0,05.24 = 1,2 (g)

    m_MgO = 5,2 - 1,2 = 4 (g)

b,\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)

⇒ n_{HCl đã dùng} = 0,1+0,2 = 0,3 (mol)

\(C_{M_{ddHCl}}=\dfrac{0,3}{0,5}=0,6\left(l\right)\)

c,\(C_{M_{MgCl_2}}=\dfrac{0,05+0,1}{0,6}=0,25M\) 

a) Gọi x là số mol Na₂CO₃

y là số mol MgCO₃

Ta có PT:

Na₂CO₃ + 2HCl ---> 2NaCl + CO₂ + H₂O

....x..............2x............2x.........x...........x

MgCO₃ + 2HCl ---> MgCl₂ + CO₂ + H₂O

....y............2y................y...........y.........y

Theo bài ra ta có:

106x + 84y = 3,96(1)

\(n_{CO_2}\)= \(\frac{0,896}{22,4}\)=0,04(mol)

<=> x+y= 0,4(2)

Giải 1 và 2 ta có:

x=0,015 ➝m_NaCl = 0,015. 58,5 = 0,8775(g)

y=0,025 ➝m_{\(MgCl_2\)}= 0,025.95=2,375(g)

m_HCl= \(\frac{3,65.100}{100}\)=3,65(g)

n_HCl = \(\frac{3,65}{36.5}\)= 0,1(mol)

=> n_{HCl dư} = 0,1 - (2x+2y) = 0,02(mol)

m_{HCl dư} = 0,02 . 36,5 = 0,73(g)

Áp dụng ĐLBTKL ta có:

m_x + m_HCl = m_Y + m_{\(CO_2\)}

=> m_y = 3,69 + 100 - 0,04.44= 101,93(g)

C%_\(NaCl\)=\(\frac{0,8775}{101,93}.100\%\)=0,86%

C%_{\(MgCl_2\)}=\(\frac{2,375}{101,93}.100\%\)=2,33%

C%_{HCl dư}= \(\frac{0,73}{101,93}.100\%\)=0,72%

b) Ta có PT

MgCl₂ + 2NaOH ---> Mg(OH)₂ + 2NaCl

...0,025.....0,05

HCl + NaOH ---> NaCl + H₂O

..0,02......0,02

=> V_{NaOH 0,35M} = \(\frac{0,05+0,02}{0,35}\)= 0,2 lít

\(n_{Zn}=\dfrac{4,55}{65}=0,07(mol)\\ Zn+2HCl\to ZnCl_2+H_2\\ a,n_{HCl}=0,14(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,14}{0,2}=0,7M\\ b,n_{H_2}=0,07(mol)\\ \Rightarrow V_{H_2}=0,07.22,4=1,568(l)\\ c,n_{ZnCl_2}=0,07(mol)\\ \Rightarrow m_{ZnCl_2}=0,07.136=9,52(g)\\ c,ZnCl_2+2AgNO_3\to 2AgCl\downarrow+Zn(NO_3)_2\)

\(m_{dd_{ZnCl_2}}=200.0,8+4,55-0,07.2=164,41(g)\\ n_{AgCl}=0,14(mol);n_{Zn(NO_3)_2}=0,07(mol)\\ \Rightarrow C\%_{Zn(NO_3)_2}=\dfrac{0,07.189}{164,41+200-0,14.143,5}.100\%=3,84%\)