Đáp án:

Bài 3:

`n_(CO_2)=1.568/22.4=0.07(mol)`

`n_(NaOH)=6.4/40=0.16(mol)`

`PTHH:2NaOH+CO_2->Na_2CO_3+H_2O`

Lập tỉ lệ:

`(n_(CO_2)/1=0.07/1=0.07<(n_(NaOH)/2=0.16/2=0.08`

`->CO_2` hết và `NaOH` dư

Theo pt: `n_(NaOH)` pứ `=2n_(CO_2)=0.07×2=0.14(mol)`

`->n_(NaOH)` dư `=0.16-0.14=0.02(mol)`

`m_(NaOH)` dư `=0.02×40=0.8(g)`

Theo pt: `n_(Na_2CO_3)=n_(CO_2)=0.07(mol)`

`m_(Na_2CO_3)=0.2×106=21.2(g)`

Bài 4:

`n_(Al)=32.4/27=1.2(mol)`

`PTHH:2Al+3CuCl_2->2AlCl_3+3Cu`

Theo pt: `n_(Cu)=3/2n_(Al)=3/2×1.2=1.8(mol)`

`m_A=m_(Cu)=1.8×64=115.2(g)`

Theo pt: `n_(CuCl_2)=n_(Cu)=1.8(mol)`

`V_(CuCl_2)=1.8/1.5=1.2(l)`

`PTHH:AlCl_3+3NaOH->Al(OH)_3+3NaCl`

`Al(OH)_3+NaOH->NaAlO_2+2H_2O`

Em tham khảo nha:

3)

\(\begin{array}{l}
a)\\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O\\
b)\\
{n_{C{O_2}}} = \dfrac{{1,568}}{{22,4}} = 0,07\,mol\\
{n_{NaOH}} = \dfrac{{6,4}}{{40}} = 0,16mol\\
\dfrac{{0,16}}{2} > \dfrac{{0,07}}{1} \Rightarrow NaOH \text{ dư}\\
{n_{NaOH}} \text{ dư}= 0,16 - 0,07 \times 2 = 0,02\,mol\\
{m_{NaOH}} \text{ dư}= 0,02 \times 40 = 0,8g\\
c)\\
{n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,07\,mol\\
{m_{N{a_2}C{O_3}}} = 0,07 \times 106 = 7,42g\\
4)\\
1.\\
2Al + 3CuC{l_2} \to 2AlC{l_3} + 3Cu\\
{n_{Al}} = \dfrac{{32,4}}{{27}} = 1,2\,mol\\
{n_{Cu}} = 1,2 \times \frac{3}{2} = 1,8\,mol\\
{m_{Cu}} = 1,8 \times 64 = 115,2g\\
2.\\
{n_{CuC{l_2}}} = 1,2 \times \dfrac{3}{2} = 1,8\,mol\\
{V_{{\rm{dd}}CuC{l_2}}} = \dfrac{{1,8}}{{1,5}} = 1,2l\\
3.\\
AlC{l_3} + 3NaOH \to Al{(OH)_3} + 3NaCl\\
Al{(OH)_3} + NaOH \to NaAl{O_2} + 2{H_2}O
\end{array}\)