Đáp án đúng: B
7,9.
Nếu NaOH (0,4 mol) và KOH (0,2 mol) phản ứng hết thì chất rắn sau khi nung 42,86 gam bao gồm NaNO2 (0,4) và KNO2 (0,2)
$\displaystyle {{\text{m}}_{\text{NaN}{{\text{O}}_{\text{2}}}}}\text{ + }{{\text{m}}_{\text{KN}{{\text{O}}_{\text{2}}}}}\text{ = 0,4}\text{.69 + 0,2}\text{.85 = 44,6 }>\text{ 42,86}$ ⇒ NaOH và KOH còn dư.
$\displaystyle \underbrace{\left\{ \begin{array}{l}\text{Fe (a)}\\\text{Cu (b) }\end{array} \right.}_{\text{14,8 (gam)}}\text{+ HN}{{\text{O}}_{\text{3}}}\text{ (0,96)}\xrightarrow{{}}\text{s}\text{.p}\text{.k + }\underbrace{{{\text{H}}_{\text{2}}}\text{O}}_{\text{0,48 mol}}\text{+ dd X}\left\{ \begin{array}{l}\text{F}{{\text{e}}^{3+}}\\\text{F}{{\text{e}}^{2+}}\text{(or }{{\text{H}}^{+}})\\\text{C}{{\text{u}}^{2+}}\\\text{NO}_{3}^{-}\text{ (y) }\end{array} \right.\text{+ }\left\{ \begin{array}{l}\text{N}{{\text{a}}^{+}}\text{ (0,4)}\\{{\text{K}}^{+}}\text{ (0,2)}\\\text{O}{{\text{H}}^{-}}\text{ ( 0,6)}\end{array} \right.\left| \begin{array}{l}\xrightarrow{{}}\text{ }\downarrow \text{ Y }\xrightarrow{{{\text{t}}^{0}}}\text{ }\underbrace{\left\{ \begin{array}{l}\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}\text{ (a/2)}\\\text{CuO (b) }\end{array} \right.}_{\text{20 (gam)}}\\\xrightarrow{{}}\text{dd Z}\left\{ \begin{array}{l}\text{N}{{\text{a}}^{+}}\text{(0,4)}\\{{\text{K}}^{+}}\text{ (0,2)}\\\text{O}{{\text{H}}^{-}}\text{ (x)}\\\text{NO}_{3}^{-}\text{ (y) }\end{array} \right.\xrightarrow{{{\text{t}}^{0}}}\underbrace{\left\{ \begin{array}{l}\text{N}{{\text{a}}^{+}}\text{(0,4)}\\{{\text{K}}^{+}}\text{(0,2)}\\\text{O}{{\text{H}}^{-}}\text{ (x)}\\\text{NO}_{2}^{-}\text{ (y)}\end{array} \right.}_{\text{42,86 (gam)}}\end{array} \right.$
Ta có hệ:
$\displaystyle \left\{ \begin{array}{l}\text{56a + 64b = 14,8}\\\text{160}\text{.a/2 + 80b = 20}\end{array} \right.\text{ }\Leftrightarrow \text{ }\left\{ \begin{array}{l}\text{a = 0,15}\\\text{b = 0,1}\end{array} \right.\text{ }$
$\displaystyle \left\{ \begin{array}{l}\text{x + y = 0,6}\\\text{17x + 46y + 23}\text{.0,4 + 39}\text{.0,2 = 42,86}\end{array} \right.\text{ }\Leftrightarrow \text{ }\left\{ \begin{array}{l}\text{x = 0,06}\\\text{y = 0,54}\end{array} \right.\text{ }$
Nếu dung dịch X gồm Fe3+ (0,15) và Cu2+ (0,1) thì
lượng$\displaystyle \text{O}{{\text{H}}^{-}}$ phản ứng = 0,15.3 + 0,1.0 = 0,65 > (0,6 – 0,06).
⇒ dd sau phản ứng có Fe2+⇒ HNO3 phản ứng hết
⇒ Trong dd X gồm có
$\displaystyle \left\{ \begin{array}{l}\text{F}{{\text{e}}^{3+}}\text{ (c)}\\\text{F}{{\text{e}}^{2+}}\text{ (d)}\\\text{C}{{\text{u}}^{2+}}\text{ (0,1)}\\\text{NO}_{3}^{-}\text{ (0,54) }\end{array} \right.$
⇒
$\displaystyle \left\{ \begin{array}{l}\text{c + d = 0,15}\\\text{3c + 2d + 0,1}\text{.2 = 0,54}\end{array} \right.\text{ }\Leftrightarrow \text{ }\left\{ \begin{array}{l}\text{c = 0,04}\\\text{d = 0,11}\end{array} \right.\text{ }$
ms.p.k = (0,96.63) – (0,48.18) – (0,54.62) = 18,36 gam
mdd sau phản ứng = 14,8 + 126 – 18,36 = 122,44 (gam)
$\displaystyle \Rightarrow \text{ C}{{{\scriptstyle{}^{\text{o}}\!\!\diagup\!\!{}_{\text{o}}\;}}_{\text{Fe(N}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{3}}}}}\text{ = }\frac{\text{0,04 }\!\!\times\!\!\text{ 242}}{\text{122,44}}\text{ }\!\!\times\!\!\text{ 100 }\approx \text{ 7,9 (}{\scriptstyle{}^{\text{o}}\!\!\diagup\!\!{}_{\text{o}}\;}\text{)}$
