To the set of the characteristics and quantitative relations of the numbers of present chemical species in a reaction of – the name of estequiometria. We must remember that to adjust a chemical equation we use the coefficients solely. In none in case that we change sub indices of formulas. If to make this we go to modify the identity of the substance. The burning of the alcohol is described for the following chemical equation. Click John Stankey to learn more.

We go to start the balancing? C2H6O + O2 Co2 + H2O As to choose the coefficients? We must start the rightness for the element that appears one alone time of each side of the equation (in this in case that we have carbon and hydrogen). Therefore, we must multiply carbon for 2 and hydrogen for 3 (both of the right side) to be with 2 carbon atoms and 6 hydrogen atoms of each side of the equation. We will have, therefore: C2H6O + O2 2CO2 + 3H2O Now we go to give one looked at for the oxignios. We have 4 pertaining oxignios to Co2 and 3 oxignios of the water, adding a total of 7 oxignios of the side of products and only 3 of the side of the reagents (1 atom of oxygen of the C2H6O and 2 atoms of the O2). How we can decide this? It is enough to multiply the oxygen for three! C2H6O + 3O2 2CO2 + 3H2O We have thus the balanced equation. Following the same reasoning when balancing the chemical equation of the burning of the gasoline we will have the following equation: C8H18 + 25/2O2 8CO2 + 9H2O We can observe that the stoichiometric coefficient of the carbon dioxide in the gasoline is greater that the alcohol, this happens, therefore the amount of hydro-carbons in the gasoline is bigger, and consequently an increase in the amount of molecule formation of carbon dioxide in the same one.