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![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_K=\dfrac{3.9}{39}=0.1\left(mol\right)\)
\(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\)
\(0.1...................0.1.....0.05\)
\(m_{H_2O}=26.2\cdot1=26.2\left(g\right)\)
\(m_{KOH}=0.1\cdot56=5.6\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=m_K+m_{H_2O}-m_{H_2}=3.9+26.2-0.05\cdot2=30\left(g\right)\)
\(C\%_{KOH}=\dfrac{5.6}{30}\cdot100\%=18.67\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{7,436}{24,79}=0,3\left(mol\right)\)
PTHH :
\(2Al+6HCL\rightarrow2AlCl_3+3H_2\uparrow\)
0,2 0,6 0,2 0,3
\(b,C_{M\left(HCl\right)}=\dfrac{0,6}{0,2}=3\left(M\right)\)
\(c,m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{FeCl_2}=\dfrac{25,4}{127}=0,2\left(mol\right)\\ V_{dd}=200ml=0,2l\\ \rightarrow C_{M\left(FeCl_2\right)}=\dfrac{0,2}{0,2}=1M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
C% CuSO4 = 16/(16 + 184) .100% = 8%
b)
n NaOH = 20/40 = 0,5(mol)
CM NaOH = 0,5/4 = 0,125M
![](https://rs.olm.vn/images/avt/0.png?1311)
250ml = 0,25 lít
\(C_{M_{KOH}}=\dfrac{0,5}{0,25}=2M\)
\(n_{KCl}=\dfrac{27.5}{74.5}=0.36\left(mol\right)\)
\(C_{M_{KCl}}=\dfrac{0.36}{0.2}=1.8\left(M\right)\)