Bài 4:

Tóm tắt:

`m_{MgCl_2}``=``19` `gam`

`m_{NaOH}``,` `m_{Mg(OH)_2}``,` `m_{NaCl}``=` `?` `gam`

Bài làm:

`n_{MgCl_2}``=``\frac{m}{M}``=``\frac{19}{95}``=``0,2` `(mol)`

`PTHH`             `MgCl_2``+``2NaOH``→``Mg(OH)_2``+``2NaCl`

                            `0,2`             `0,4`                `0,2`               `0,4`        `mol`

`→``n_{NaOH}``=``n_{MgCl_2}``=``\frac{0,2.2}{1}``=``0,4` `(mol)`

     `m_{NaOH}``=``n``.``M``=``0,4``.``40``=``16` `(g)`

`→``n_{Mg(OH)_2}``=``n_{MgCl_2}``=``0,2` `(mol)`

     `m_{Mg(OH)_2}``=``n``.``M``=``0,2``.``58``=``11,6` `(g)`

`→``n_{NaCl}``=``n_{MgCl_2}``=``\frac{0,2.2}{1}``=``0,4` `(mol)`

     `m_{NaCl}``=``n``.``M``=``0,4``.``58,5``=``23,4` `(g)`

Bài 5:

Tóm tắt:

`m_{CuO}``=``4` `gam`

`V_{H_2}``=``?` `l`

`m_{Cu}``,` `m_{H_2O}``=``?` `gam`

Bài làm:

`n_{CuO}``=``\frac{m}{M}``=``\frac{4}{80}``=``0,05` `(mol)`

`PTHH`               `CuO``+``H_2` $\xrightarrow[]{t^o}$ `Cu``+``H_2O`

                            `0,05`     `0,05`      `0,05`     `0,05`       `mol`

`→``n_{H_2}``=``n_{CuO}``=``0,05` `(mol)`

     `V_{H_2}``=``n``.``22,4``=``0,05``.``22,4``=``1,12` `(l)`

`→``n_{Cu}``=``n_{CuO}``=``0,05` `(mol)`

     `m_{Cu}``=``n``.``M``=``0,05``.``64``=``3,2` `(g)`

`→``n_{H_2O}``=``n_{CuO}``=``0,05` `(mol)`

     `m_{H_2O}``=``n``.``M``=``0,05``.``18``=``0,9` `(g)`

Câu 4:

Tóm tắt: 

$m_{MgCl_2}=19(g)$

Tìm $m_{Mg(OH)_2};m_{NaOH};m_{NaCl}$

Giải:

$n_{MgCl_2}=\dfrac{19}{90}=0,2(mol)$

$PTHH:MgCl_2+2NaOH\to Mg(OH)_2\downarrow+2NaCl$

Theo PT: $n_{MgCl_2}=n_{Mg(OH)_2}=0,2(mol);n_{NaOH}=n_{NaCl}=0,4(mol)$

$\Rightarrow m_{Mg(OH)_2}=0,2.58=11,6(g)$

$m_{NaOH}=0,4.40=16(g)$

$m_{NaCl}=0,4.58,5=23,4(g)$

Bài 5:

Tóm tắt:

$m_{CuO}=4(g)$

Tìm $V_{H_2(đktc)};m_{Cu};m_{H_2O}$

Giải:

$n_{CuO}=\dfrac{4}{80}=0,05(mol)$

$PTHH:CuO+H_2\xrightarrow{t^o}Cu+H_2O$

Theo PT: $n_{Cu}=n_{H_2}=n_{H_2O}=n_{CuO}=0,05(mol)$

$\Rightarrow V_{H_2(đktc)}=0,05.22,4=1,12(lít)$

$m_{Cu}=0,,05.64=3,2(g)$

$m_{H_2O}=0,05.18=0,9(g)$