a.
| 2Fe2O3 | + 3C | → 4Fe | + 3CO2 | |
| \[m\] | ①\[25\mathrm{kg}\] | ④\[2,8\mathrm{kg}\] | ||
| \[n\] | ②\[156,543519\mathrm{mol}\] | ③\[234,815279\mathrm{mol}\] |
① \(m_\mathrm{Fe_2O_3} = 25\mathrm{kg} = 25 \cdot 10^3\mathrm{g} = 25000\mathrm{g}\)
② \(n_\mathrm{Fe_2O_3} = \frac {m_\mathrm{Fe_2O_3}}{M_\mathrm{Fe_2O_3}} = \frac {25000\mathrm{g}}{(55,85 \cdot 2 + 16,00 \cdot 3)\frac {\mathrm{g}}{\mathrm{mol}}} = 156,543519\mathrm{mol}\)
③
\(n_\mathrm{Fe_2O_3}:n_\mathrm{C} = 2:3 \\ n_\mathrm{C} = \frac {3}{2}n_\mathrm{Fe_2O_3} = \frac {3}{2} \cdot 156,543519\mathrm{mol} = 234,815279\mathrm{mol}\)
④ \(m_\mathrm{C} = 12,01\frac {\mathrm{g}}{\mathrm{mol}} \cdot 234,815279\mathrm{mol} = 2820,13150\mathrm{g} = 2,82013150 \cdot 10^3\mathrm{g} ≈ 2,8\mathrm{kg}\)
b.
| 2Fe2O3 | + 3C | → 4Fe | + 3CO2 | |
| \[m\] | ①\[500\mathrm{kg}\] | ④\[350\mathrm{kg}\] | ||
| \[n\] | ②\[3130,87038\mathrm{mol}\] | ③\[6261,74076\mathrm{mol}\] |
① \(m_\mathrm{Fe_2O_3} = 500\mathrm{kg} = 500 \cdot 10^3\mathrm{g}\)
② \(n_\mathrm{Fe_2O_3} = \frac {m_\mathrm{Fe_2O_3}}{M_\mathrm{Fe_2O_3}} = \frac {500 \cdot 10^3\mathrm{g}}{(55,85 \cdot 2 + 16,00 \cdot 3)\frac {\mathrm{g}}{\mathrm{mol}}} = 3130,87038\mathrm{mol}\)
③
\(n_\mathrm{Fe_2O_3}:n_\mathrm{Fe} = 1:2 \\ n_\mathrm{Fe} = 2n_\mathrm{Fe_2O_3} = 2 \cdot 3130,87038\mathrm{mol} = 6261,74076\mathrm{mol}\)
④
\[\begin{aligned} m_\mathrm{Fe} &= M_\mathrm{Fe} \cdot n_\mathrm{Fe} = 55,85\frac {\mathrm{g}}{\mathrm{mol}} \cdot 6261,74076\mathrm{mol} = 349718,222\mathrm{g} = \\ &= 349,718222 \cdot 10^3\mathrm{g} ≈ 350\mathrm{kg}\end{aligned}\]
c.
| 2Fe2O3 | + 3C | → 4Fe | + 3CO2 | |
| \[m\] | ①\[572\mathrm{ton}\] | ④\[338\mathrm{ton}\] | ||
| \[n\] | ②\[1,02417189 \cdot 10^7\mathrm{mol}\] | ③\[7,68128917 \cdot 10^6\mathrm{mol}\] |
① \(m_\mathrm{Fe} = 572\mathrm{ton} = 572 \cdot 10^6\mathrm{g}\)
② \(n_\mathrm{Fe} = \frac {m_\mathrm{Fe}}{M_\mathrm{Fe}} = \frac {572 \cdot 10^6\mathrm{g}}{55,85\frac {\mathrm{g}}{\mathrm{mol}}} = 1,02417189 \cdot 10^7\mathrm{mol}\)
③
\(n_\mathrm{Fe}:n_\mathrm{CO_2} = 4:3 \\ n_\mathrm{CO_2} = \frac {3}{4}n_\mathrm{Fe} = \frac {3}{4} \cdot 1,02417189 \cdot 10^7\mathrm{mol} = 7,68128917 \cdot 10^6\mathrm{mol}\)
④
\(\begin{aligned}m_\mathrm{CO_2} &= M_\mathrm{CO_2} \cdot n_\mathrm{CO_2} = (12,01 + 16,00 \cdot 2)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 7,68128917 \cdot 10^6\mathrm{mol} = \\ &= 338,053536 \cdot 10^6\mathrm{g} ≈ 338\mathrm{ton}\end{aligned}\)
