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Giải thích các bước giải:

Câu 9:

`a) Fe + 2HCl —> FeCl_2 + H_2`

`b)`

`n_{Fe}=4,2÷56=0,075 mol`

Theo pt: `n_{FeCl_2}=n_{Fe}=0,075 mol`

`=> m_{FeCl_2} = 0,075×(56+35,3×2)=9,525g`

`c)`

Theo pt: `n_{H_2}=n_{Fe}=0,075 mol`

`=> V_{H_2}=0,075×22,4=1,68l`

Câu 10:

`a)`

_`FeCl_2: `

`M_{FeCl_2}=56+35,3×2=127` g/mol

`=> %Fe=56÷127×100=44,09%`

     `%Cl=100%-44,09%=55,91%`

_`HCl:`

`M_{HCl}=1+35,3=36,5` g/mol
`=> %H=1÷36,5×100=2,74%`
     `%Cl=100%-2,74%=97,26%`

`b)`

_`NaOH:`

`M_{NaOH}=23+16+1=40` g/mol

`=> %Na=23÷40×100=57,5%`

     `%O=16÷40×100=40%`

     `%H=100%-57,5%-40%=2,5%`

_`KNO_3:`

`M_{KNO_3}=39+14+16×3=101` g/mol
`=> %K=39÷101×100=38,6%`
     `%N=14÷101×100=13,86%`
     `%O=100%-38,6%-13,86%=47,54%`

Câu 9.

a/ PTHH: $Fe+2HCl\to FeCl_2+H_2$

b/ $n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{4,2}{56}=0,075(mol)$

Theo pt: $n_{FeCl_2}=0,075(mol)$

$\to m_{FeCl_2}=n_{FeCl_2}.M_{FeCl_2}=0,075.127=9,525(g)$

c/ Theo pt: $n_{H_2}=0,075(mol)$

$\to V_{H_2}=n_{H_2}.22,4=0,075.22,4=1,68(l)$

Bài 10.

a/ Xét nguyên tố $FeCl_2$

$\%Fe=\dfrac{M_{Fe}.100}{M_{FeCl_2}}\%=44\%$

$\%Cl=\dfrac{M_{Cl}.2.100}{M_{FeCl_2}}\%=56\%$

Xét nguyên tố $HCl$

$\%H=\dfrac{M_H.100}{M_{HCl}}\%=3\%$

$\%Cl=\dfrac{M_{Cl}.100}{M_{HCl}}\%=97\%$

b/ Xét nguyên tố $NaOH$:

$\%Na=\dfrac{M_{Na}.100}{M_{NaOH}}\%=57,5\%$

$\%O=\dfrac{M_O.100}{M_{NaOH}}\%=40\%$

$\%H=\dfrac{M_H.100}{M_{NaOH}}\%=2,5\%$

Xét nguyên tố $KNO_3$:

$\%K=\dfrac{M_K.100}{M_{KNO_3}}\%=38\%$

$\%N=\dfrac{M_N.100}{M_{KNO_3}}\%=14\%$

$\%O=\dfrac{M_O.3.100}{M_{KNO_3}}\%=48\%$