3.19

a)
$H_2O\left(g\right)+Cl_2O\left(g\right)\xrightleftharpoons[]{}2\ HOCl\left(g\right)$

$K_c=0{,}090$

lasketaan kaasujen ainemäärät

$n\left(H_2O\right)=\frac{1{,}0g}{18{,}01528\ \frac{g}{mol}}=0{,}05550...mol\approx55{,}5mmol$
$n\left(Cl_2O\right)=\frac{2{,}0g}{86{,}9054\ \frac{g}{mol}}=0{,}02301...mol\approx23{,}0mmol$

$K_c=\frac{\left[HOCl\right]^2}{\left[H_2O\right]\left[Cl_2O\right]}\Rightarrow0{,}090=\frac{x^2}{\left(0{,}0555-x\right)\left(0{,}02301-x\right)}$
$\left(x=−0{,}015772\right)\ tai\ x=0{,}008007$
$\begin{array}{l|l} &H_2O&Cl_2O&HOCl\\ \hline c_{alku}&0{,}0555\ mol&0{,}0230\ mol&0\\ muutos&-0{,}008007&-0{,}008007&+0{,}008007\\ c_{tasap}&0{,}04749&0{,}015003&0{,}008007 \end{array}$
$\left[H_2O\right]=0{,}047mol$
$\left[Cl_2O\right]=0{,}015mol$
$\left[HOCl\right]=0{,}0080mol\$

b)

lasketaan HOCl-kaasun ainemäärä

$n\left(HOCl\right)=\frac{100g}{52{,}46\ \frac{g}{mol}}=1{,}90621...mol$
$K_c=0{,}090$
$K_c=\frac{\left[HOCl\right]^2}{\left[H_2O\right]\left[Cl_2O\right]}\Rightarrow0{,}090=\frac{\left(1{,}90621...-x\right)^2}{x^2}$
$x=1.46631\ \left(tai\ x=2.72315\right)$

$\begin{array}{l|l} &2\ HOCl&H_2O&Cl_2O\\ \hline c_{alku}&1{,}90621&0&0\\ muutos&-x&+\frac{1}{2}x&+\frac{1}{2}x\\ c_{tasap}&0{,}4399...&0{,}7331&0{,}7331 \end{array}$
$\left[HOCl\right]=0{,}440\ mol$
$\left[H_2O\right]=0{,}733\ mol$
$\left[Cl_2O\right]=0{,}733\ mol$