Đáp án:
$H_{x}I_{y}O_{z}+\frac{m(2z-x)}{2(n-\frac{5}{x}m)}HNO_{3}→\frac{2z-x}{2(n-\frac{5}{x}m)}N_{m}O_{n}+\frac{y}{2}I_{2}+[\frac{x}{2}+\frac{m(2z-x)}{4(n-\frac{5}{x}m)}]H_{2}O$
Giải thích các bước giải:
Đặt $aH_{x}I_{y}O_{z}+ bHNO_{3} -→ cN_{m}O_{n}+ dI_{2}+ eH_{2}O$
$⇒ \begin{cases}ax+b=2e(*)\\ay=2d\\az+3b=cn+e(**)\\b=cm\end{cases}$
$\text{Cho $a=1$ ⇒ $d=\dfrac{y}{2}$}$
$(*)⇔ x+b=2e$
$(**)⇔ z+3b=cn+e$
$⇔ 2z+6b=2cn+2e$
$⇔ 2z+6cm=2cn+x+cm$
$⇔ 2cn-5cm=2z-x$
$⇔ 2c(n-\dfrac{5}{2}m)=2z-x$
$⇔ c=\dfrac{2z-x}{2(n-\dfrac{5}{x}m)}$
$⇒ b=cm=\dfrac{m(2z-x)}{2(n-\dfrac{5}{x}m)}$
$⇒ e=\dfrac{x+b}{2}$
$=\dfrac{x+\dfrac{m(2z-x)}{2(n-\dfrac{5}{x}m)}}{2}$
$=\dfrac{\dfrac{2x(n-\dfrac{5}{x}m)+m(2z-x)}{2(n-\dfrac{5}{x}m)}}{2}$
$=\dfrac{2x(n-\dfrac{5}{x}m)+m(2z-x)}{4(n-\dfrac{5}{x}m)}$
$=\dfrac{x}{2}+\dfrac{m(2z-x)}{4(n-\dfrac{5}{x}m)}$
$\text{Phương trình khi được cân bằng là:}$
$H_{x}I_{y}O_{z}+\frac{m(2z-x)}{2(n-\frac{5}{x}m)}HNO_{3}→\frac{2z-x}{2(n-\frac{5}{x}m)}N_{m}O_{n}+\frac{y}{2}I_{2}+[\frac{x}{2}+\frac{m(2z-x)}{4(n-\frac{5}{x}m)}]H_{2}O$
