Đáp án đúng: B
0,12.
Dung dịch Y có Fe3+, H+ dư, Cl– (1,16), $\text{NO}_{\text{3}}^{\text{-}}$
Từ Y + NaOH → 0,84.1,5 = 3.0,38 + nH+ dư → nH+ dư = 0,12
Y + Fe → nFe3+ + 3NO = 2nFe ⇒ nFe3+ = 0,4. BT điện tích ⇒${{\text{n}}_{\text{NO}_{\text{3}}^{\text{-}}}}\text{=0,16}$
$\displaystyle \text{X(30,4}\,\text{gam)}\left\{ \begin{array}{l}\text{F}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}\,\,\text{(x)}\\\text{FeC}{{\text{O}}_{\text{3}}}\,\,\text{(y)}\\\text{Fe(OH}{{\text{)}}_{\text{2}}}\,\text{(z)}\end{array} \right.\text{+}\left\{ \begin{array}{l}\text{Fe(N}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{3}}}\,\,\text{(t)}\\\text{HCl}\,\text{(1,16)}\end{array} \right.$$\displaystyle \xrightarrow{{}}\text{Y}\left\{ \begin{array}{l}\text{F}{{\text{e}}^{\text{3+}}}\,\text{(0,4)}\\{{\text{H}}^{\text{+}}}\,\text{(0,12)}\\\text{C}{{\text{l}}^{\text{-}}}\,\text{(1,16)}\\\text{NO}_{\text{3}}^{\text{-}}\,\text{(0,16)}\end{array} \right.\text{+}\left\{ \begin{array}{l}\text{C}{{\text{O}}_{\text{2}}}\,\text{(y)}\\\text{NO}\,\text{(3a-0,16)}\end{array} \right.\text{+}{{\text{H}}_{\text{2}}}\text{O}$
BT e có x + y + z = 3.(3t=0,16)
BT Fe có 3x + y + z + t = 0,4 ⇒ 2x + 3(3t-0,16) + t = 0,4 (1)
$\displaystyle \text{X(30,4}\,\text{gam)}\left\{ \begin{array}{l}\text{F}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}\,\,\text{(x)}\\\text{FeC}{{\text{O}}_{\text{3}}}\,\,\text{(y)}\\\text{Fe(OH}{{\text{)}}_{\text{2}}}\,\text{(z)}\end{array} \right.\text{+HCl}\xrightarrow{{}}\text{(43,48}\,\text{gam)}\left\{ \begin{array}{l}\text{FeC}{{\text{l}}_{\text{2}}}\,\text{(x+y+z}\,\text{=}\,\text{3(3t-0,16))}\\\text{FeC}{{\text{l}}_{\text{3}}}\,\,\text{(2x)}\end{array} \right.$
⇒ 127.3.(3t-0,16) + 162,5.2x = 43,48 (2)
Từ (1), (2) ⇒ x = 0,04; t = 0,08 ⇒ y + z = 0,2
Từ khối lượng hỗn hợp X ⇒ 116y + 90z = 21,12
⇒ y = 0,12; z = 0,08
