Cho m gam SO3 vào 200g dung dịch H2SO4 14,7% thu được dung dịch H2SO4 20%.Viết PTHH của phản ứng,tính m
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![](https://rs.olm.vn/images/avt/0.png?1311)
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\(SO_3+H_2O\rightarrow H_2SO_4\\ n_{SO_3}=a\left(mol\right)\\ \rightarrow m_{SO_3}=80a\left(g\right);m_{H_2SO_4}=98a\left(g\right)\\ Vì:dd.thu.được.nồng.độ.20\%,nên.ta.có:\\ \dfrac{200.14,7\%+98a}{80a+200}.100\%=20\%\\ \Leftrightarrow a=12,927\\ Vậy:m=m_{SO_3}=12,927\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2SO_4}=\dfrac{200\cdot19.6\%}{98}=0.4\left(mol\right)\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(0.4......................0.4\)
\(m_{SO_3}=0.4\cdot80=32\left(g\right)\)
\(b.\)
\(n_{H_2SO_4}=\dfrac{80\cdot19.6\%}{98}=0.16\left(mol\right)\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
\(0.16..........0.16..............0.16\)
\(m_{MgO}=0.16\cdot40=6.4\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=6.4+80=86.4\left(g\right)\)
\(C\%MgSO_4=\dfrac{0.16\cdot120}{86.4}\cdot100\%=22.22\%\)
a)
$SO_3 + H_2O \to H_2SO_4$
n SO3 = n H2SO4 = 200.19,6%/98 = 0,4(mol)
=> m = 0,4.80 = 32(gam)
b)
$MgO + H_2SO_4 \to MgSO_4 + H_2O$
n MgSO4 = n MgO = n H2SO4 = 80.19,6%/98 = 0,16(mol)
=> m MgO = 0,16.40 = 6,4(gam)
Sau pư, m dd = 6,4 + 80 = 86,4(gam)
=> C% MgSO4 = 0,16.120/86,4 .100% = 22,22%
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ b.n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\\ n_{H_2SO_4}=3.0,15=0,45\left(mol\right)\\ m_{ddH_2SO_4}=\dfrac{0,45.98.100}{14,7}=300\left(g\right)\\ m_{ddsau}=24+300=324\left(g\right)\\ n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,15\left(mol\right)\\ C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{0,15.400}{324}.100\approx18,519\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n_{SO_3}=\dfrac{m}{M}=\dfrac{32}{80}=0,4\left(mol\right)\)
PTHH: `SO_3 + H_2O -> H_2SO_4`
b) Theo PTHH: `n_{H_2SO_4} = n_{SO_3} = 0,4 (mol)`
`=> m_{H_2SO_4} = 0,4.98 = 39,2 (g)`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4}=n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\\ \Rightarrow m_{CuSO_4}=0,1\cdot160=16\left(g\right)\\ b,n_{H_2SO_4}=n_{CuO}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{H_2SO_4}}=0,1\cdot98=9,8\left(g\right)\\ \Rightarrow C\%_{H_2SO_4}=\dfrac{9,8}{200}\cdot100\%=4,9\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(SO_3+H_2O\rightarrow H_2SO_4\\ n_{H_2SO_4}=n_{SO_3}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ a,m=m_{H_2SO_4}=98.0,3=29,4\left(g\right)\\ b,C\%_{ddH_2SO_4}=\dfrac{29,4}{0,3.80+150}.100\approx16,897\%\\ c,H_2SO_4:Tính.axit\Rightarrow Quỳ.tím.hoá.đỏ\)
\(SO_3+H_2O\rightarrow H_2SO_4\\ m_{H_2SO_4\left(tăng\right)}=\dfrac{98}{80}m=\dfrac{49}{40}m=1,225m\left(g\right)\\ m_{H_2SO_4\left(dd.14,7\%\right)}=14,7.200=29,4\left(g\right)\\ Ta.có:C\%_{ddH_2SO_4\left(cuối\right)}=20\%\\ \Leftrightarrow\dfrac{29,4+1,225m}{m+200}.100\%=20\%\\ m\approx10,3415\left(g\right)\)