Zusammenfassung der Ressource
Chemical calculations
- Percentage
composition
- %of C in CO2 = total relative mass of C in CO2 *100
- relative molecular mass of CO2
- Mr=12+(2*16)
- Example-CO2
- = 12/44*100= 12.3%
- % of O in CO2 =
- total relative mass of O in CO2
- relative molecular mass of CO2
- =32/44*100= 72.7%
- Mass of reactants and products
- Example-Na2CO3+2HLC------->H2O+CO2
- Relative mass Sodium Carbonate=106
- Relative molecular mass of carbon dioxide=44
- 106g of Na2CO3 forms 44g of CO2
- 1g of Na2CO3 forms 44/106 of CO2
- 5.3g of Na CO3 forms 44/106*5.3 of CO2
- = 2.2g
- Yield of a chemical reaction