Količina tvari

Pr. 1

Izračunajte množinu tvari klora u 200 g.

${\displaystyle m(Cl_{2})=200g}$

${\displaystyle n(Cl_{2})=?}$

${\displaystyle n(Cl_{2})={\frac {m(Cl_{2})}{M(Cl_{2})}}={\frac {200~g}{(2\cdot 35,45)\cdot gmol^{-1}}}={\frac {200~g}{70,9~gmol^{-1}}}=2,82~mol}$

Pr. 2

Izračunaj množinu iona natrija u 0,5 g kuhinjske soli.

${\displaystyle m(NaCl)=0,5g}$

${\displaystyle n(Na^{+})=?}$

${\displaystyle n(NaCl)={\frac {m(NaCl)}{M(NaCl)}}={\frac {0,5~g}{(22,99+35,45)\cdot gmol^{-1}}}={\frac {0,5~g}{58,44~gmol^{-1}}}=8,56\cdot 10^{-3}~mol}$

${\displaystyle NaCl\longrightarrow Na^{+}+Cl^{-}\qquad \Rightarrow \qquad {\frac {n(NaCl)}{n(Na^{+})}}={\frac {1}{1}}\qquad \Rightarrow \qquad n(Na^{+})=n(NaCl)}$

${\displaystyle {\begin{alignedat}{2}n(Na^{+})&=n(NaCl)\\&=8,56\cdot 10^{-3}~mol\\\end{alignedat}}}$

Pr. 3

U uzorku kromova(III) klorida mase 0,08 g izračunaj množinu kloridnih iona.

${\displaystyle m(CrCl_{3})=0,08g}$

${\displaystyle n(Cl^{-})=?}$

${\displaystyle n(CrCl_{3})={\frac {m(CrCl_{3})}{M(CrCl_{3})}}={\frac {0,08~g}{(52,00+3\cdot 35,45)\cdot gmol^{-1}}}={\frac {0,08~g}{158,35~gmol^{-1}}}=5,05\cdot 10^{-4}~mol}$

${\displaystyle CrCl_{3}\longrightarrow Cr^{3+}+3~Cl^{-}\qquad \Rightarrow \qquad {\frac {n(CrCl_{3})}{n(Cl^{-})}}={\frac {1}{3}}\qquad \Rightarrow \qquad n(Cl^{-})=3\cdot n(CrCl_{3})}$

${\displaystyle {\begin{alignedat}{2}n(Cl^{-})&=3\cdot n(CrCl_{3})\\&=3\cdot 5,05\cdot 10^{-4}~mol\\&=1,52\cdot 10^{-3}~mol\\\end{alignedat}}}$

• Avogadrov zakon

1. Primjeri 2 i 3 djelomično su preuzeti iz:

A. Petreski - B. Sever, Zbirka riješenih primjera i zadataka iz opće kemije, PROFIL international, 4. izdanje, Zagreb, 1997.