PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
Ta có: \(n_{MgO}=\dfrac{10}{40}=0,25\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,5\left(mol\right)\\n_{MgCl_2}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{0,5}{0,5}=1\left(M\right)\\C_{M_{MgCl_2}}=\dfrac{0,25}{0,5}=0,5\left(M\right)\end{matrix}\right.\)
Ta có: nMgO = \(\dfrac{10}{40}=0,25\left(mol\right)\)
a. PTHH: MgO + 2HCl ---> MgCl2 + H2O
Theo PT: nHCl = 2.nMgO = 2.0,25 = 0,5(mol)
Đổi 500ml = 0,5 lít
=> \(C_{M_{HCl}}=\dfrac{0,25}{0,5}=0,5M\)
b. Ta có: \(V_{dd_{MgCl_2}}=10+0,5=10,5\left(lít\right)\)
Theo PT: \(n_{MgCl_2}=n_{MgO}=0,25\left(mol\right)\)
=> \(C_{M_{MgCl_2}}=\dfrac{0,25}{10,5}=0,024M\)
1. K2O + SO2 ---> K2SO3.
2. K2O + N2O5 ---> 2KNO3.
3. K2O + CO2 ---> K2CO3.
4. BaO + SO2 ---> BaSO3.
5. BaO + N2O5 ---> N2O + BaO5.
6. BaO + CO2 ---> BaCO3.
7. CuO + CO2 ---> CuCO3.
Ta có: \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
a. PTHH: Al2O3 + 6HCl ---> 2AlCl3 + 3H2O
Theo PT: \(n_{HCl}=6.n_{Al_2O_3}=6.0,1=0,6\left(mol\right)\)
=> mHCl = 0,6 . 36,5 = 21,9(g)
=> C%HCl = \(\dfrac{21,9}{250}.100\%=8,76\%\)
b. Ta có: \(m_{dd_{AlCl_3}}=10,2+250=160,2\left(g\right)\)
Theo PT: \(n_{AlCl_3}=2.n_{Al_2O_3}=2.0,1=0,2\left(mol\right)\)
\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
=> \(C\%_{AlCl_3}=\dfrac{26,7}{160}.100\%=16,69\%\)
