Important rules for resolving a reaction calculation: | |
1 | Use balanced reaction equation and add the phases. |
2 | Underline the data, the given data and those asked for. The other substances (without any data) are not needed in the calculation.
Sometimes the data are directly given, sometimes indirectly. The calculation is continues with only the underlined substances. |
3 | write down the mol proportion |
4 | Where needed, convert the mol into the right units (to be found in the data and in the required answer) |
5 | Introduce a conversion factor to arrive at the real amounts, given in the data.
That's how you finish the calculation. |
Calculate mass and volume (standaerd conditions for gases) of carbon dioxyde produced in the complete combustion of 4,01 g methane? | |
1 | CH_{4}(g) + 2 O_{2}(g) CO_{2}(g) + 2H_{2}O(g) |
2 | Underline the substances with data. CH_{4}(g) + 2 O_{2}(g) CO_{2}(g) + 2H_{2}O(g) |
3 | So, 1 mol CH_{4}(g) reacts with 1 mol CO_{2}(g) (proportion 1:1) |
4 | 16 gramme CH_{4}(g) produce 44 gramme CO_{2}(g) (here we apply the molecular masses) |
5 | in reality we do not have 16 grammes, but only 4 grammes for combustion.
To be introduced a factor of 4/16. (in this case we can simply divide by 4) 4/16 x 16 gramme CH_{4}(g) produce 4/16 x 44 gramme CO_{2}(g) to finish: standard condition mean: at temp = 25^{o}C and pressure of 1 atm. Then 1 mol gas = 22,4 liter 1/16 x 44 = 11 gram CO_{2}(g) is produced, that equals 4/16 mol = 4/16 x 22,4 liter CO_{2}(g) = 5,6 liter |