$a,$

$m_{H_2SO_4}=\dfrac{300×4,9}{100}=14,7(g)$

$n_{H_2SO_4}=\dfrac{14,7}{98}=0,15(mol)$

$Mg+H_2SO_4→MgSO_4+H_2$

Theo PT: $n_{Mg}=n_{H_2}=n_{MgSO_4}=n_{H_2SO_4}=0,15(mol)$

$→m=0,15×24=3,6(g)$

$→V=0,15×22,4=3,36(l)$

$b,$

$→m_{MgSO_4}=0,15×120=18(g)$

$c,$

$m_{dd}=3,6+300-0,15×2=303,3(g)$

$→C\%_{MgSO_4}=\dfrac{18}{303,3}×100\%=5,93\%$

Đáp án+ Giải thích các bước giải:

`PT : Mg + H_2SO_4 -> MgSO_4 + H_2`

`n_{H_2SO_4}= {300.4,9%}/{100.98}=0,15(mol)`

`a)`

Theo pt : `n_{Mg}=n_{H_2SO_4}=0,15(mol)`

`-> m_{Mg}=n.M=0,15.24=3,6(g)=m`

Theo pt : `n_{H_2}=n_{H_2SO_4}=0,15(mol)`

`->V_{H_2}=n.22,4=0,15.22,4=3,36(l)=V`

`b)`

Theo pt : `n_{MgSO_4}=n_{Mg}=0,15(mol)`

`-> m_{MgSO_4}=n.M=0,15.120=18(g)`

`c)`

$m_{\text{dd spứ}}$ `=m_{Mg}+m_{H_2SO_4}-m_{H_2}=3,6+300-0,15.2=303,3(g)`

`-> C%`$_{\text{dd $MgSO_4$}}$ `=\frac{m_{ct}.100%}{m_{dd}}=\frac{18.100%}{303,3}\approx5,9%`

$\text{#Cyrus}$