$\text{n$_{Fe}$=28:56=0,5(mol)}$

$\text{PTHH:}$

$\text{Fe+H$_{2}$SO$_{4}$→FeSO$_{4}$+H$_{2}$↑}$

$\text{0,5→0,5    →0,5→0,5(mol)}$

$\text{a,Từ PTHH ta suy ra : n$_{H2}$=0,5(mol)}$

$\text{⇒V$_{H_2(đkc)}$=24,79.0,5=12,395(lít)}$

$\text{b,Từ PTHH ta suy ra : n$_{H2SO4}$=0,5(mol)}$

$\text{⇒m$_{H_2SO_4}$=0,5.98=49(g)}$

$\text{⇒m$_{dung dịch H2SO4}$=49:9,8%=500(g)}$

$\text{c,Từ PTHH ta suy ra : n$_{FeSO4}$=0,5(mol)}$

$\text{⇒m$_{FeSO4}$=0,5.152=76(g)}$

$\text{m$_{dung dịch sau p/ứ}$=28+500-0,5.2=527(g)}$

$\text{⇒C%$_{FeSO4}$=$\frac{76}{527}$.100% ≈14,42%}$ 

Đáp án:

`↓`

Giải thích các bước giải:

`a.` `n_{Fe}``=``\frac{m}{M}``=``\frac{28}{56}``=``0,5` `(mol)`

`PTHH`               `Fe``+``H_2SO_4``→``FeSO_4``+``H_2↑`

                            `0,5`        `0,5`               `0,5`          `0,5`           `mol`

`→``n_{H_2}``=``n_{Fe}``=``0,5` `(mol)`

`→`$V_{H_2(đkc)}$`=``n``.``24,79``=``0,5``.``24,79``=``12,395` `(l)`

`b.`Theo pt, ta có: `n_{H_2SO_4}``=``n_{Fe}``=``0,5` `(mol)`

`→``m_{H_2SO_4}``=``n``.``M``=``0,5``.``98``=``49` `(g)`

`→`$m_{dd_{H_2SO_4}}$`=``\frac{m_{H_2SO_4}.100%}{C%}``=``\frac{49.100%}{9,8%}``=``500` `(g)`

`c.` `m_{H_2}``=``n``.``M``=``0,5``.``2``=``1` `(g)`

     `m_{sau}``=``m_{Fe}``+`$m_{dd_{H_2SO_4}}$`-``m_{H_2}``=``28``+``500``-``1``=``527` `(g)`

Theo pt, ta có: `n_{FeSO_4}``=``n_{Fe}``=``0,5` `(mol)`

`→``m_{FeSO_4}``=``n``.``M``=``0,5``.``152``=``76` `(g)`

`→``C%_{FeSO_4}``=``\frac{m_{FeSO_4}.100%}{m_{sau}}``=``\frac{76.100%}{527}``≈``14,42%`