| CaCO3 | → CaO | + CO2 | |
| \[m\] | ④\[178,477175\mathrm{g}\] | ①\[100\mathrm{g}\] | |
| \[n\] | ③\[1,78316690\mathrm{mol}\] | ②\[1,78316690\mathrm{mol}\] |
① \(m_\mathrm{CaO} = 100\mathrm{g}\)
② \(n_\mathrm{CaO} = \frac {m_\mathrm{CaO}}{M_\mathrm{CaO}} = \frac {100\mathrm{g}}{(40,08+16,00)\frac {\mathrm{g}}{\mathrm{mol}}} = 1,78316690\mathrm{mol}\)
③ \(n_\mathrm{CaCO_3}:n_\mathrm{CaO} = 1:1 \Rightarrow n_\mathrm{CaCO_3} = n_\mathrm{CaO} = 1,78316690\mathrm{mol}\)
④
\(\begin{aligned} m_\mathrm{CaCO_3} &= M_\mathrm{CaCO_3} \cdot n_\mathrm{CaCO_3} = \\ &= (40,08 + 12,01 + 16,00 \cdot 3)\mathrm{g/mol} \cdot 1,78316690\mathrm{mol} = \\ &= 178,477175\mathrm{g}\end{aligned}\)
Den massa CaCO3 som faktist behövs beräknas:
\[m_\mathrm{CaCO_3,behov} = \frac {m_\mathrm{CaCO_3}}{\text{relativt utbyte}} = \frac {178,477175\mathrm{g}}{0,79} = 225,920475\mathrm{g} ≈ 0,23\mathrm{kg}\]
Svar: Det behövs 0,23kg kalciumkarbonat.
