Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
\(2NaOH+SO_2-->Na_2SO_3+H_2O\left(1\right)\)
0,2________0,1__________0,1
\(n_{SO_2}=0,1\left(mol\right)\)
a) \(m_{Na_2SO_3}=0,1.126=12,6\left(g\right)\)
b) \(m_{d^2_{NaOH}}=\frac{0,2.40.100}{25}=32\left(g\right)\)
c) \(2NaOH+H_2SO_4-->Na_2SO_4+2H_2O\left(2\right)\)
0,2___________0,1
=> \(V_{d^2H_2SO_4}=\frac{0,1.98.100}{20.1,14}=42,98\left(ml\right)\)
\(n_{H_2S}=\dfrac{0.448}{22.4}=0.02\left(mol\right)\)
\(n_{NaOH}=0.1\cdot0.5=0.05\left(mol\right)\)
\(T=\dfrac{0.05}{0.02}=2.5>2\)
\(2NaOH+H_2S\rightarrow Na_2S+H_2O\)
\(0.04........0.02..............0.02\)
\(n_{Na_2S}=0.02\left(mol\right)\)
\(n_{NaOH\left(dư\right)}=0.05-0.04=0.01\left(mol\right)\)
\(n_{NaOH}=0.24\cdot0.1=0.024\left(mol\right)\)
\(T=\dfrac{0.024}{0.02}=1.2\)
=> Tạo 2 muối
\(n_{Na_2S}=a\left(mol\right),n_{NaHS}=b\left(mol\right)\)
\(\left\{{}\begin{matrix}2a+b=0.024\\a+b=0.02\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=0.004\\b=0.016\end{matrix}\right.\)
a, \(2Al+3S\rightarrow Al_2S_3\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(Al_2S_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2S\)
\(n_X=0,15\left(mol\right)\)
\(\overline{M_X}=27,6\)
Gọi \(\left\{{}\begin{matrix}n_{H2}:a\left(mol\right)\\n_{H2S}:b\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=0,15\\\frac{2a+34b}{0,15}=27,6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,03\\b=0,12\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Al\left(dư\right)}=0,02\left(mol\right)\\n_{Al2S3}=0,04\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Al\left(pư\right)}=0,08\left(mol\right)\\n_{S\left(pư\right)}=0,12\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow a=m_{Al}+m_S=27.\left(0,02+0,08\right)+32.0,12=6,54\left(g\right)\)
b,\(2H_2+O_2\rightarrow2H_2O\)
0,03__________0,03
\(H_2S+\frac{3}{2}O_2\rightarrow SO_2+H_2O\)
0,12____________012__
\(n_{KOH}=\frac{112.10\%}{56}=0,2\left(mol\right)\)
\(\frac{n_{KOH}}{n_{SO2}}=1,6\Rightarrow\) Tạo muối \(\left\{{}\begin{matrix}K_2SO_3:x\left(mol\right)\\KHSO_3:y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x+y=0,2\\x+y=0,12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,08\\y=0,04\end{matrix}\right.\)
\(m_{dd\left(pư\right)}=m_{SO2}+m_{H2O}+112=120,22\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{K2SO3}=10,51\%\\C\%_{KHSO3}=4\%\end{matrix}\right.\)
c,\(n_{SO2}=\frac{3n_{Al}+4n_S}{2}=0,39\left(mol\right)\)
\(n_{Ca\left(OH\right)2}=0,5\left(mol\right)\Rightarrow\) Dư kiềm
\(Ca\left(OH\right)_2+SO_2\rightarrow CaSO_3+H_2O\)
0,39_____039______0,39___
\(\Rightarrow m_{\downarrow}=46,8\left(g\right)\)
n N a O H = 1.2= 2 mol
⇒ Tạo ra 2 muối NaHS và N a 2 S
⇒ Chọn C.