Kpl.2.4

2.18
a)

$MgCO_3\left(s\right)\rightarrow MgO\left(g\right)+CO_2\left(g\right)$

b)
$2KClO_3\left(s\right)\rightarrow2KCl\left(g\right)+3O_2\left(g\right)$

c)
$2H_2O_2\left(s\right)\rightarrow2H_2O\left(l\right)+O_2\left(g\right)$
d)
$NH_4NO_3\left(s\right)\rightarrow N_2O\left(g\right)+2H_2O\left(g\right)$
$2N_2O\left(g\right)\rightarrow2N_2\left(g\right)+O_2\left(g\right)$
e)

$O_3\left(g\right)\rightarrow O_2\left(g\right)+O\left(g\right)$
f)
$2AgF\left(s\right)\rightarrow2Ag\left(s\right)+F_2\left(g\right)$

2.19
a)

$2NaN_3\rightarrow3N_2\left(g\right)+2Na\left(l\right)$

b)
$V\left(N_2\right)=40l$

$\rho\left(N_2\right)=1{,}20\ \frac{g}{l}$
$m\left(N_2\right)=1{,}20g\cdot40l=48g$

$m\left(NaN_3\right)=?$

$n\left(N_2\right)=\frac{m}{M}=\frac{48{,}00\ g}{28{,}02\ \frac{g}{mol}}=1{,}713\ mol$

$\frac{n\left(NaN_3\right)}{n\left(N_3\right)}=\frac{2}{3}$ ,joten

$n\left(NaN_3\right)=\frac{2}{3}\cdot1{,}713mol=1{,}142\ mol$

$m\left(NaN_3\right)=nM\cdot1{,}142\ mol\cdot65{,}02\ \frac{g}{mol}=74{,}25\approx74\ g$

2.22
$n\left(kaasut\right)=1{,}56\ mol$
$M\left(C_2H_4N_2O_6\right)=152{,}072\ \frac{g}{mol}$
$M\left(H_2O\right)=18{,}016\ \frac{g}{mol}$
$m\left(C_2H_4N_2O_6\right)=?$
$m\left(H_2O\right)=?$

$C_2H_4N_2O_6\rightarrow2CO_2\left(g\right)+2H_2O\left(g\right)+N_2\left(g\right)$

$\frac{n\left(C_2H_4N_2O_6\right)}{n\left(kaasut\right)}=\frac{1}{5}$ , joten

$n\left(C_2H_4N_2O_6\right)=\frac{1}{5}\cdot1{,}56\ mol=0{,}312\ mol$

$m\left(C_2H_4N_2O_6\right)=nM=0{,}312\ mol\cdot152{,}072\ \frac{g}{mol}=47{,}446464\approx47{,}4\ g$

$\frac{n\left(H_2O\right)}{n\left(C_2H_4N_2O_6\right)}=\frac{2}{1}$ , joten

$n\left(H_2O\right)=2\cdot0{,}312\ mol=0{,}624\ mol$

$m\left(H_2O\right)=nM=0{,}624\ mol\cdot18{,}016\ \frac{g}{mol}=11{,}241984...\approx11{,}2g$