Trộn 200ml dung dịch H2SO4 24,5% ( d = 1,12 g/ml ) vào 26 gam dung dịch H2SO4 10%. Tính C% của H2SO4 trong dung dịch thu được sau pha trộn.
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
n H 2 SO 4 = 0 , 2 x 2 , 5 + 0 , 1 x 1 = 0 , 6 ( mol )
→ C M sau khi trộn = 0,6/0,3 = 2M.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(V_{\text{dd}}=200+250=450ml=0,45l\\ n_{H_2SO_4}=\left(0,2.1\right)+\left(2.0,25\right)=0,7\left(mol\right)\\ C_M=\dfrac{0,7}{0,45}=1,5M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
PT ion: \(H^++OH^-\rightarrow H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{OH^-}=0,6\cdot0,4+0,6\cdot0,3\cdot2=0,6\left(mol\right)\\n_{H^+}=0,2\cdot2,6=0,52\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\) H+ hết, OH- còn dư \(\Rightarrow n_{OH^-\left(dư\right)}=0,08\left(mol\right)\)
\(\Rightarrow\left[OH^-\right]=\dfrac{0,08}{0,6+0,2}=0,1\left(M\right)\) \(\Rightarrow pH=14+log\left(0,1\right)=13\)
Bài 2:
PT ion: \(H^++OH^-\rightarrow H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{OH^-}=0,3\cdot1,6=0,48\left(mol\right)\\n_{H^+}=0,2\cdot1\cdot2+0,2\cdot2=0,8\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\) OH- hết, H+ còn dư \(\Rightarrow n_{H^+\left(dư\right)}=0,32\left(mol\right)\)
\(\Rightarrow\left[H^+\right]=\dfrac{0,32}{0,2+0,3}=0,64\left(M\right)\) \(\Rightarrow pH=-log\left(0,64\right)\approx0,19\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: n H 2 SO 4 = 0 , 2 x 2 , 5 + 0 , 1 x 1 = 0 , 6 ( mol )
→ C M sau khi trộn = 0 , 6 / 0 , 3 = 2 M .
![](https://rs.olm.vn/images/avt/0.png?1311)
1. \(n_{H_2SO_4\left(98\%\right)}=\dfrac{30.1,84.98\%}{98}=0,552\left(mol\right)\)
=>\(V_{H_2SO_4\left(1M\right)}=\dfrac{0,552}{1}=0,552\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(m_{ddH_2SO_4\left(60\%\right)}=700.1,503=1052,1\left(g\right)\Rightarrow m_{H_2SO_4}=1052,1.60\%=631,26\left(g\right)\)
\(m_{ddH_2SO_4\left(20\%\right)}=500.1,1476=573,8\left(g\right)\Rightarrow m_{H_2SO_4}=573,8.20\%=114,76\left(g\right)\)
ΣmH2SO4 = 631,26 + 114,76 = 746,02 (g)
\(n_{H_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)
PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,08\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,08.98=7,84\left(g\right)\)
\(\Rightarrow\dfrac{746,02}{V}=\dfrac{7,84}{0,2}\Rightarrow V\approx19,03\left(l\right)\)
\(m_{ddH_2SO_4.24,5\%}=200\times1,12=224\left(g\right)\)
\(\Rightarrow m_{H_2SO_4.24,5\%}=224\times24,5\%=54,88\left(g\right)\)
\(m_{H_2SO_4.10\%}=26\times10\%=2,6\left(g\right)\)
Ta có: \(m_{ddH_2SO_4}mới=224+26=250\left(g\right)\)
\(m_{H_2SO_4}mới=54,88+2,6=57,48\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}mới=\frac{57,48}{250}\times100\%=22,992\%\)