Câu 3:
a. Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
=> \(V_{H_2}=0,2.22,4=4,48\left(lít\right)\)
b. Theo PT: \(n_{HCl}=2.n_{Zn}=2.0,2=0,4\left(mol\right)\)
=> \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
Ta có: \(C_{\%_{HCl}}=\dfrac{14,6}{m_{dd_{HCl}}}.100\%=10\%\)
=> \(m_{dd_{HCl}}=146\left(g\right)\)
c. Ta có: \(m_{dd_{ZnCl_2}}=13+146-\left(0,2.2\right)=158,6\left(g\right)\)
Theo PT: \(n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\)
=> \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
=> \(C_{\%_{ZnCl_2}}=\dfrac{27,2}{158,6}.100\%=17,15\%\)
Câu 4:
a.
(1) Na2O + H2O ---> 2NaOH
(2) 2NaOH + H2SO4 ---> Na2SO4 + 2H2O
(3) Na2SO4 + 2KCl ---> K2SO4 + 2NaCl
(4) 2NaCl + Ba(NO3)2 ---> BaCl2 + 2NaNO3
b.
(1) \(2Fe\left(OH\right)_3\overset{t^o}{--->}Fe_2O_3+3H_2O\)
(2) Fe2O3 + 6HCl ---> 2FeCl3 + 3H2O
(3) 2FeCl3 + 3Ba(NO3)2 ---> 2Fe(NO3)3 + 3BaCl2
(4) Fe(NO3)3 + 3NaOH ---> Fe(OH)3 + 3NaNO3
(5) 2Fe(OH)3 + 3H2SO4 ---> Fe2(SO4)3 + 3H2O
