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\(n_{H_2S}=\dfrac{0.448}{22.4}=0.02\left(mol\right)\)
\(n_{NaOH}=0.1\cdot0.5=0.05\left(mol\right)\)
\(T=\dfrac{0.05}{0.02}=2.5>2\)
\(2NaOH+H_2S\rightarrow Na_2S+H_2O\)
\(0.04........0.02..............0.02\)
\(n_{Na_2S}=0.02\left(mol\right)\)
\(n_{NaOH\left(dư\right)}=0.05-0.04=0.01\left(mol\right)\)
\(n_{NaOH}=0.24\cdot0.1=0.024\left(mol\right)\)
\(T=\dfrac{0.024}{0.02}=1.2\)
=> Tạo 2 muối
\(n_{Na_2S}=a\left(mol\right),n_{NaHS}=b\left(mol\right)\)
\(\left\{{}\begin{matrix}2a+b=0.024\\a+b=0.02\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=0.004\\b=0.016\end{matrix}\right.\)
nNaOH = 0,17.1 = 0,17 (mol)
Xét tỉ lệ: \(\dfrac{n_{NaOH}}{n_{H_2S}}=\dfrac{0,17}{0,1}=1,7\)
=> Tạo muối NaHS và Na2S
=> A
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
\(2NaOH+H_2S\rightarrow Na_2S+2H_2O\)
0,035_____0,0175___0,0175_____ \(Na_2S+H_2S\rightarrow2NaHS\) 0,0025___0,0025___0,005 \(\Rightarrow\left\{{}\begin{matrix}CM_{Na2S}=\frac{0,0175-0,0025}{0,1}=0,15M\\CM_{NaHS}=\frac{0,005}{0,1}=0,05M\end{matrix}\right.\)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
\(n_{ZnS}=0,125\left(mol\right)\)
\(n_{Zn\left(Dư\right)}=0,2-0,125=0,075\left(mol\right)\)
\(ZnS+2HCl\rightarrow ZnCl_2+H_2S\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(\left\{{}\begin{matrix}n_{H2S}=n_{ZnS}=0,125\left(mol\right)\\n_{H2}=n_{Zn}=0,075\left(mol\right)\end{matrix}\right.\)
\(M_{hh}=\frac{0,125.34+0,075.2}{0,2}=22\)
\(D_{hh/H2}=\frac{22}{2}=11\)
1) \(n_{H_2S}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{NaOH}=0,2.1=0,2\left(mol\right)\)
Xét \(\dfrac{n_{NaOH}}{n_{H_2S}}=\dfrac{0,2}{0,1}=2\) => Tạo muối Na2S
PTHH: 2NaOH + H2S --> Na2S + 2H2O
0,2------------>0,1
=> mNa2S = 0,1.78 = 7,8 (g)
2)
\(n_{SO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{NaOH}=0,2.1=0,2\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{n_{NaOH}}{n_{SO_2}}=\dfrac{0,2}{0,1}=2\) => Tạo muối Na2SO3
PTHH: 2NaOH + SO2 --> Na2SO3 + H2O
0,2-------------->0,1
=> mNa2SO3 = 0,1.126 = 12,6 (g)
2. \(\left\{{}\begin{matrix}n_{SO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\n_{NaOH}=0,2.1=0,2\left(mol\right)\end{matrix}\right.\)
Ta có: \(T=\dfrac{0,2}{0,1}=2\) ⇒ tạo ra muối Na2SO3
SO2 + 2NaOH -----> Na2SO3 + H2O
nH2S = V/22.4 = 8.96/22.4 = 0.4 (mol)
nNaOH = CM.V = 0.5 x 1 = 0.5 (mol)
Ta có: 1<nNaOH / nH2S = 0.5/0.4 = 5/4 = 1.25 <2
==> tạo ra 2 muối
NaOH + H2S => NaHS + H2O
2NaOH + H2S => Na2S + H2O
Chọn C