Kpl 3.2

3.8

$m\left(AgNO_3+NaCl\right)=3{,}012g$

$m\left(AgCl\right)=2{,}466g$

$M\left(AgCl\right)=107{,}87\frac{g}{mol}+35{,}45\ \frac{g}{mol}=143{,}32\ \frac{g}{mol}$

$M\left(AgNO_3\right)=107{,}87\ \frac{g}{mol}+14{,}01\ \frac{g}{mol}+3\cdot16{,}00\ \frac{g}{mol}=169{,}88\ \frac{g}{mol}$

$n\left(AgCl\right)=\frac{m\left(AgCl\right)}{M\left(AgCl\right)}=\frac{2{,}466\ g}{143{,}32\ \frac{g}{mol}}=0{,}01720625...\approx0{,}0172063\ mol$

$Ag^+\left(aq\right)+Cl^-\left(aq\right)\rightarrow AgCl\left(s\right)$

Saostumisreaktion reaktioyhtälön perusteella

$n\left(Ag^+\right)=n\left(AgCl\right)$

$n\left(AgNO_3\right)=n\left(AgCl\right)$

$m\left(AgNO_3\right)=nM=0{,}0172063\ mol\cdot169{,}88\ \frac{g}{mol}=2{,}923006...\approx2{,}92301g$

$m-\%\left(AgNO_3\right)=\frac{2{,}92301\ g}{3{,}012\ g}=0{,}97045...\approx0{,}9705=97{,}05\%$

Näin ollen natriumnitraatin massaprosentti on

$m-\%\left(NaNO_3\right)=100\%-97{,}05\%=2{,}95\%$

3.10

$Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O$

$m\left(Na_2CO_3+NaCl\right)=3{,}00g$

$V\left(CO_2\right)=315cm^3=315\cdot10^{-6}m^3$

$p=96{,}5\ kPa=96{,}5\cdot10^3\ Pa$

$R=8{,}31451\ \frac{Pa\cdot m^3}{mol\cdot K}$

$T\left(21{,}0+273{,}15\right)K=294{,}15K$

$M\left(Na_2CO_3\right)=105{,}99\ \frac{g}{mol}$

$pV=nRT\ \leftrightarrow\ n=\frac{pV}{RT}$
$n\left(CO_2\right)=\frac{96{,}5\cdot10^3\ Pa\cdot315\cdot10^{-6}m^3}{8{,}31451\ \frac{Pa\cdot m^3}{mol\cdot K}\cdot294{,}15K}=0{,}01242889...\approx0{,}0124289\ mol$

a)-kohdan reaktioyhtälön perusteella

$n\left(Na_2CO_3\right)=n\left(CO_2\right)=0{,}0124289\ mol$

$m\left(NaCO_3\right)=nM=0{,}0124289\ mol\cdot105{,}99\ \frac{g}{mol}=1{,}31733...\approx1{,}3173\ g$
$m-\%\left(NaCO_3\right)=\frac{1{,}3173\ g}{3{,}00\ g}=0{,}4391=43{,}91\%$

3.11

Titraus 1:

$EDTA^{4-}\ reagoi\ sekä\ Mg^{2+}{,}\ että\ Ca^{2+}ionien\ kanssa$

$n\left(EDTA^{4-}\right)=c\cdot V=0{,}0104\ \frac{mol}{l}\cdot0{,}0315\ l=3{,}276\cdot10^{-4}mol=n\left(Mg^{2+}\right)+n\left(Ca^{2+}\right)$

Titraus 2:

$EDTA^{4-}\ reagoi\ vain\ Mg^{2+}\ ionien\ kanssa\ \left(Ca^{2+}\ ionit\ on\ saostettu\ sulfatteina\right)$
$n\left(EDTA^{4-}\right)=c\cdot V=0{,}0104\ \frac{mol}{l}\cdot0{,}0187\ l=1{,}9448\cdot10^{-4}\ mol=n\left(Mg^{2+}\right)$

$c\left(Mg^{2+}\right)=\frac{1{,}9448\cdot10^{-4}\ mol}{0{,}010\ l}=1{,}9448\cdot10^{-3}\ mol\approx1{,}94\cdot10^{-3}\ mol$

$c\left(Ca^{2+}\right)=\frac{1{,}3312\cdot10^{-4}\ mol}{0{,}010\ l}=1{,}3312\cdot10^{-3}\ \frac{mol}{l}\approx1{,}33\cdot10^{-3}\ \frac{mol}{l}$

3.12

$2Al\left(s\right)+3H_2SO_4\left(aq\right)\rightarrow Al_2\left(SO_4\right)_3\left(aq\right)+3H_2\left(g\right)$

$Zn\left(s\right)+H_2SO_4\left(aq\right)\rightarrow ZnSO_4\left(aq\right)+H_2\left(g\right)$

$m\left(Zn+Al\right)=1{,}00g$

$m\left(H_2\right)=84{,}0\ mg=84{,}0\cdot10^{-3}g$

$M\left(H_2\right)=2{,}016\ \frac{g}{mol}$
$n\left(H_2\right)=\frac{m}{M}=\frac{84{,}0\cdot10^{-3}\ g}{2{,}016\ \frac{g}{mol}}=0{,}0416666..\approx0{,}041667\ mol$

$n\left(Zn\right)=n\left(H_2\right)=0{,}04167\ mol$

$\frac{n\left(H_2\right)}{n\left(Al\right)}=\frac{3}{2}$ , joten $n\left(H_{_2}\right)=\frac{3}{2}n\left(Al\right)$

$n\left(H_2\right)=n\left(Zn\right)+\frac{3}{2}n\left(Al\right)$

Oletetaan, että alumiinin maassa on x ja sinkin 1,00-x

$0{,}04167=\frac{3}{2}\cdot\frac{x\ g}{26{,}98\ \frac{g}{mol}}+\frac{\left(1{,}00-x\right)g}{65{,}38\ \frac{g}{mol}}=0{,}654437\ g\approx0{,}654\ g$

3.13

$Al(OH)_3+Mg\left(OH\right)_2$

$Al(OH)_3\left(aq\right)+3HCl\left(aq\right)\rightarrow AlCl_3\left(aq\right)+3H_2O\left(l\right)$

$Mg\left(OH\right)_2\left(aq\right)+2HCl\left(aq\right)\rightarrow MgCl_2\left(aq\right)+2H_2O\left(l\right)$

$V\left(HCl\right)=37{,}7\ ml=37{,}7\cdot10^{-3}l=37{,}7\cdot10^{-3}dm^3$

$c\left(HCl\right)=2{,}50\ \frac{mol}{dm^3}$

$m\left(Al(OH)_3+Mg\left(OH\right)_2\right)=2{,}50g$

$M\left(Al\left(OH\right)_4\right)=78{,}004\ \frac{g}{mol}$
$M\left(Mg\left(OH\right)_2\right)=58{,}326\ \frac{g}{mol}$

$n\left(HCl\right)=c\cdot V=2{,}50\ \frac{mol}{dm^3}\cdot37{,}7\cdot10^{-3}dm^3=0{,}09425\ mol$

$\frac{n\left(HCl\right)}{n\left(Al\left(OH\right)_3\right)}=\frac{3}{1}$ , joten $n\left(HCl\right)=3\cdot n\left(Al(OH)_3\right)$

$\frac{n\left(HCl\right)}{n\left(Mg\left(OH\right)_2\right)}=\frac{2}{1}$ , joten $n\left(HCl\right)=2\cdot n\left(Mg\left(OH\right)_2\right)$

$n\left(HCl\right)=3\cdot n\left(Al(OH)_3\right)+2\cdot n\left(Mg\left(OH\right)_2\right)$

Oletetaan, että Alumiinihydraatin massa on x ja magnesiumhydraatin 2,50-x $0{,}09425\ mol=3\cdot\left(\frac{x\ g}{78{,}004\ \frac{g}{mol}}\right)+2\cdot\left(\frac{\left(2{,}50-x\right)\ g}{58{,}326\ \frac{g}{mol}}\right)$
$m\left(Al(OH)_3\right)=x\approx2.04\ g$
$m\left(Mg\left(OH\right)_2\right)=2.5-x\approx0{,}455g$