Kpl.1.3

1.12
a) Oikein
b) Väärin, 0,67 mol
c) Oikein
d) Väärin, 4,0 mol
e) Oikein
f) Oikein

1.13
a)
$2H_2\left(g\right)+O_2\left(g\right)\rightarrow2H_2O\left(l\right)$
$m\left(H\right)=10{,}0g$
$M\left(H_2\right)=1{,}008\frac{g}{mol}\cdot2=2{,}016\frac{g}{mol}$

$n\left(H_2\right)=\frac{m}{M}=\frac{10{,}0g}{2{,}016\frac{g}{mol}}=4{,}96031...\approx4{,}9603mol$

$\frac{n\left(O_2\right)}{n\left(H_2\right)}=\frac{1}{2}$
$n\left(O_2\right)=\frac{1}{2}\cdot n\left(H_2\right)=\frac{1}{2}\cdot4{,}9603\frac{g}{mol}=2{,}048015mol$

$M\left(O_2\right)=16{,}00\frac{g}{mol}\cdot2=32{,}00\frac{g}{mol}$

$m\left(O_2\right)=n\cdot M=2{,}048015mol\cdot32{,}00\frac{g}{mol}=79{,}3648...\approx76{,}4g$
b)

$2H_2\left(g\right)+O_2\left(g\right)\rightarrow2H_2O\left(l\right)$
$m\left(H\right)=10{,}0g$
$M\left(H_2\right)=1{,}008\frac{g}{mol}\cdot2=2{,}016\frac{g}{mol}$

$n\left(H_2\right)=\frac{m}{M}=\frac{10{,}0g}{2{,}016\frac{g}{mol}}=4{,}96031...\approx4{,}9603mol$

$\frac{n\left(H_2O\right)}{n\left(H_2\right)}=\frac{2}{2}$
$n\left(H_2O\right)=\frac{2}{2}\cdot n\left(H_2\right)=1\cdot4{,}9603\ mol=4{,}9603mol$

$M\left(H_2O\right)=18{,}01528\frac{g}{mol}$

$m\left(H_2O\right)=n\cdot M=4{,}9603\frac{g}{mol}\cdot18{,}01528\frac{g}{mol}=89{,}3611...\approx89{,}4g=0{,}0894kg$
c)

$2H_2\left(g\right)+O_2\left(g\right)\rightarrow2H_2O\left(l\right)$
H=4
O=2

1.14
a)
$N_2H_4\left(g\right)+O_2\left(g\right)\rightarrow N_2\left(g\right)+2H_2O\left(g\right)$
b)
$N_2H_4\left(g\right)+O_2\left(g\right)\rightarrow N_2\left(g\right)+2H_2O\left(g\right)$
$m\left(N_2H_4\right)=150g$
$M\left(N_2H_4\right)=32{,}0452\frac{g}{mol}$

$n\left(N_2H_4\right)=\frac{m}{M}=\frac{150{,}0g}{32{,}0452\frac{g}{mol}}=4{,}6808...\approx4{,}681mol$

$\frac{n\left(O_2\right)}{n\left(N_2H_4\right)}=\frac{1}{1}$
$n\left(O_2\right)=\frac{1}{1}\cdot n\left(N_2H_4\right)=1\cdot4{,}681mol=4{,}681mol$

$M\left(O_2\right)=2\cdot16{,}00=32{,}00\frac{g}{mol}$

$m\left(O_2\right)=n\cdot M=4{,}681mol\cdot32{,}00\frac{g}{mol}=149{,}792\approx150g$
c)

$N_2H_4\left(g\right)+O_2\left(g\right)\rightarrow N_2\left(g\right)+2H_2O\left(g\right)$
$m\left(N_2H_4\right)=3{,}0kg=3000g$
$M\left(N_2H_4\right)=32{,}0452\frac{g}{mol}$

$n\left(N_2H_4\right)=\frac{m}{M}=\frac{3000g}{32{,}0452\frac{g}{mol}}=93{,}6177...\approx93{,}618mol$

$\frac{n\left(kaasut\right)}{n\left(N_2H_4\right)}=\frac{3}{1}$

$n\left(kaasut\right)=3\cdot93{,}618mol=280{,}853...\approx280mol$

1.15
a)
$2Na_2O_2\left(g\right)+2H_2O\left(g\right)\rightarrow O_2\left(g\right)+4NaOH\left(aq\right)$

$m\left(O_2\right)=0{,}50g$
$M\left(O_2\right)=2\cdot16{,}00\frac{g}{mol}=32{,}00\frac{g}{mol}$

$n\left(O_2\right)=\frac{m}{M}=\frac{0{,}50g}{32{,}0452\frac{g}{mol}}=0{,}015625mol$

$\frac{n\left(Na_2O_2\right)}{n\left(O_2\right)}=\frac{2}{1}=2$

$n\left(Na_2O_2\right)=2\cdot0{,}015625mol=0{,}03125mol$

$M\left(Na_2O_2\right)=77{,}98\frac{g}{mol}$

$m\left(O_2\right)=n\cdot M=0{,}03125mol\cdot77{,}98\frac{g}{mol}=2{,}436875g\approx2{,}4g$
b)

$2Na_2O_2\left(g\right)+2H_2O\left(g\right)\rightarrow O_2\left(g\right)+4NaOH\left(aq\right)$

$V\left(NaOH\right)=75ml=0{,}075dm^3$
$M\left(O_2\right)=2\cdot16{,}00\frac{g}{mol}=32{,}00\frac{g}{mol}$

$n\left(O_2\right)=\frac{m}{M}=\frac{0{,}50g}{32{,}0452\frac{g}{mol}}=0{,}015625mol$

$\frac{n\left(NaOH\right)}{n\left(O_2\right)}=\frac{4}{1}$
$n\left(NaOH\right)=n\left(O_2\right)\cdot4=0{,}0625mol$
$c=\frac{n}{V}=\frac{0{,}0625mol}{0{,}075dm^3}=0{,}83333...\approx0{,}83\frac{mol}{l}$

1.16

$2KClO_3\left(s\right)\rightarrow3O_2\left(g\right)+2KCl\left(s\right)$

$m\left(KClO_3\right)=2{,}00g$
$m\left(Ruisku\right)=11{,}450g$
$m\left(Ruisku+O_2\right)=12{,}170g$

$m\left(O_2\right)=12{,}170g-11{,}450g=0{,}72g$

$M\left(KClO_3\right)=122{,}55\frac{g}{mol}$

$n\left(KClO_3\right)=\frac{2{,}00g}{122{,}55\frac{g}{mol}}=0{,}01631986944\approx0{,}016320mol$

$\frac{n\left(O_2\right)}{n\left(KClO_3\right)}=\frac{3}{2}$

$n\left(O_2\right)=\frac{3}{2}\cdot n\left(KClO_3\right)=0{,}02448mol$

$m\left(O_2\right)=n\cdot M=0{,}02448mol\cdot32{,}00\frac{g}{mol}=0{,}78336$

$saanto-\%\left(O_2\right)=\frac{0{,}72g}{0{,}78336...g}=0{,}9191...\approx0{,}919=91{,}9\%$

1.17
a)

$NaOH(aq)+HCl(aq)\rightarrow NaCl(aq)+H_2O(l)$

$V\left(NaOH\right)=25{,}0ml=0{,}025dm^3$
$c\left(HCl\right)=0{,}10\frac{mol}{dm^3}$

$V\left(HCl\right)=15{,}8ml=0{,}0158dm^3$

$c\left(NaOH\right)=?$

$n\left(HCl\right)=cV=0{,}10\frac{mol}{dm^3}\cdot0{,}0158dm^3=0{,}00158mol$

$\frac{n\left(HCl\right)}{n\left(NaOH\right)}=\frac{1}{1}=1$
$n\left(NaOH\right)=0{,}00158mol\cdot1=0{,}00158mol$

$c\left(NaOH\right)=\frac{n}{V}=\frac{0{,}00158mol}{0{,}025dm^3}=0{,}0632\frac{mol}{dm^3}$
b)

$Ba\left(OH\right)_2(aq)+2HNO_3(aq)\rightarrow Ba\left(NO_3\right)_2(aq)+2H_2O(l)$

$V\left(Ba(OH)_2\right)=10{,}0ml=0{,}010dm^3$
$c\left(Ba\left(OH\right)_2\right)=0{,}00250\frac{mol}{dm^3}$

$c\left(HNO_3\right)=0{,}500\frac{mol}{dm^3}$
$V\left(HNO_3\right)=?$

$n\left(Ba(OH)_2\right)=cV=0{,}00250\frac{mol}{dm^3}\cdot0{,}010dm^3=0{,}000025mol$

$\frac{n\left(HNO_3\right)}{n\left(Ba\left(OH\right)_2\right)}=\frac{2}{1}=2$
$n\left(HNO_3\right)=0{,}000025mol\cdot2=0{,}00005mol$

$V\left(NaOH\right)=\frac{n}{c}=\frac{0{,}00005mol}{0{,}500\frac{mol}{dm^3}}=0{,}0001dm^3=0{,}100ml$