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a)\(n_{CaSO_3\downarrow}=0,95\left(mol\right)\)
Bảo toàn ntố S: \(n_{SO2}=n_{CaSo_3}=0,95\left(mol\right)\)
Đặt \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Mg}=y\end{matrix}\right.\)\(\Rightarrow56x+24y=28,8\)(1)
Bảo toàn electron: \(3x+2y=2.n_{SO_2}=1,9\)
\(\Rightarrow\left\{{}\begin{matrix}56x+24y=28,8\\3x+2y=1,9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,3\\y=0,5\end{matrix}\right.\)
\(\Rightarrow\%Fe=\dfrac{56.0,3}{28,8}.100\%=58,3\%\)\(\Rightarrow\%Mg=41,7\%\)
b) Bảo toàn Mg: \(n_{Mg}=n_{MgSO_4}=0,5\left(mol\right)\) ; Bảo toàn Fe: \(n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Fe}=0,15\left(mol\right)\)
Bảo toàn S : \(n_{H_2SO_4}=n_{MgSO_4}+3n_{Fe_2\left(SO_4\right)_3}+n_{SO_2}=1,9\left(mol\right)\)
\(\Rightarrow C\%H_2SO_4=\dfrac{1,9.98}{285}.100\%=65,3\%\)
\(m_{dd}=m_{Kl}+m_{ddH_2SO_4}-m_{SO_2}=28,8+285-0,95.64=253\left(g\right)\)
(sau p/u)
\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,5.120}{253}.100\%=23,71\%\)
\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,15.400}{253}.100\%=23,71\%\)
a,\(m_{H_2SO_4}=49\%.200=98\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{98}{98}=1\left(mol\right)\)
PTHH: Zn + H2SO4 → ZnSO4 + H2
Mol: 1 1 1 1
⇒ m = 1.65 = 65 (g)
b, \(V_{H_2}=1.24=24\left(l\right)\)
c, \(m_{ZnSO_4}=1.161=161\left(g\right)\)
mdd sau pứ = 65+200-1.2=263 (g)
\(\Rightarrow C\%_{ddZnSO_4}=\dfrac{161.100\%}{263}=61,22\%\)
Câu 6:
Ta có:
\(\%m_S=\frac{32\left(n+1\right)}{98+80n}=36,7\%\)
\(\Leftrightarrow n=1,5\)
Vậy oleum này có dạng H2SO4.1,5SO3
Câu 10:
Hàm lượng S trong oleum trên là \(\%m_S=\frac{32\left(n+1\right)}{98+80n}=37,21\%\Leftrightarrow n=2\)
Oleum là H2SO4.2SO3
Ta có:
\(n_{oleum}=\frac{25,8}{98+80.2}=0,1\left(mol\right)\)
\(\Rightarrow n_{H2SO4}=3n_{oleum}=0,3\left(mol\right)\Rightarrow m_{H2SO4}=0,3.98=29,4\left(g\right)\)
\(m_{dd\left(X\right)}=25,8+74,2=100\left(g\right)\)
\(\Rightarrow C\%_{H2SO4}=\frac{29,4}{100}.100\%=29,4\%\)
Câu 12:
Gọi CTTQ của oleum là H2SO4.nSO3
\(\%_{SO3}=\frac{80n}{98+80n}.100\%=55,05\%\)
\(\Leftrightarrow n=1,5\)
\(n_{oleum}=\frac{21,8}{218}=0,1\left(mol\right)\)
\(PTHH:H_2SO_4.1,5SO_3+1,5H_2O\rightarrow2,5H_2SO_4\)
\(\Rightarrow n_{H2SO4}=n_{oleum}.2,5=0,1.2,5=0,25\left(mol\right)\)
\(m_{dd\left(spu\right)}=21,8+103,2=125\left(g\right)\)
\(\Rightarrow C\%_{H2SO4}=\frac{0,25.98}{125}.100\%=19,6\%\)
\(BaCl2+H2SO4-->BaSO4+2HCl\)
\(n_{BaSO4}=\frac{46,6}{233}=0,2\left(mol\right)\)
\(nH2SO4=n_{BaSO4}=0,2\left(mol\right)\)
\(m_{H2SO4}=0,2.98=19,6\left(g\right)\)
\(C\%_{H2SO4}=\frac{19,6}{200}.100\%=9,8\%\)
\(n_{HCl}=2n_{BaSO4}=0,4\left(mol\right)\)
\(HCl+NaOH-->NaCl+H2O\)
\(n_{NaOH}=1,8.0,5=0,9\left(mol\right)\)
\(n_{HCl}=n_{NaOH}=0,9\left(mol\right)\)
\(n_{HCl\left(X\right)}=0,9-0,4=0,5\left(mol\right)\)
\(m_{HCl\left(X\right)}=0,5.36,5=18,25\left(g\right)\)
\(C\%_{HCl}=\frac{18,25}{200}.100\%=9,125\%\)
\(m_{dung\ dịch\ sau\ pư} = m_{oleum} + m_{dd\ H_2SO_4} = 38,7 +100 = 138,7(gam)\)
\(n_{oleum} = \dfrac{38,7}{258} = 0,15(mol)\\ \Rightarrow m_{H_2SO_4\ trong\ X} = 0,15.98 + 100.30\% = 44,7(gam)\\ \Rightarrow C\%_{H_2SO_4} = \dfrac{44,7}{138,7}.100\% = 32,23\%\)
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