5)

Phản ứng xảy ra:

\(CuC{l_2} + 2NaOH\xrightarrow{{}}Cu{(OH)_2} + 2NaCl\)

Ta có:

\({n_{NaOH}} = \frac{{20}}{{40}} = 0,5{\text{ mol > 2}}{{\text{n}}_{CuC{l_2}}}\)

Vậy \(NaOH\) dư

\( \to {n_{Cu{{(OH)}_2}}} = {n_{CuC{l_2}}} = 0,2{\text{ mol}}\)

\( \to {m_{Cu{{(OH)}_2}}} = 0,2.(64 + 17.2) = 19,6{\text{ gam}}\)

7)

PHản ứng xảy ra:

\(Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{{\text{O}}_3} + {H_2}O\)

Ta có:

\({n_{C{O_2}}} = \frac{{1,56}}{{22,4}} = 0,0696{\text{ mol = }}{{\text{n}}_{CaC{{\text{O}}_3}}}\)

\( \to {m_{CaC{{\text{O}}_3}}} = 0,0696.100 = 6,96{\text{ gam}}\)

5 . 

`2NaOH + CuCl_2→ Cu(OH)3_2+ 2NaCl` 

`n_{NaOH} = {20}/{40} = 0,5(mol)`

`n_{NaOH}>2n_{CuCl_2}`

`→ NaOH` dư 

`n_{Cu(OH)_2}= n_{CuCl_2} = 0,2(mol)`

`m_{Cu(OH)_2} = 0,2.  98 = 19,6(g)` 

7. 

`CO_2 + Ca(OH)_2 → CaCO_3+ H_2O`

`n_{CaCO_3} = n_{CO_2} = {1,56}/{22,4}= 0,07(mol)`

`m = 0,07. 100 = 7(g)`