Answer
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Hint:While balancing a chemical equation we have to keep in mind that the number of atoms on each side should be equal. We can multiply the stoichiometric coefficient of each reactant and product and make each type of atom the same. Here, we have an organic compound which is propane is doing combustion in presence of oxygen and gives carbon dioxide and water.
Complete step-by-step answer:We have propane which reacts with oxygen and performs a combustion reaction. The products of combustion reaction are carbon dioxide and water. So let’s start with counting the number of atoms left hand side and right hand side. We have equation as,
$C{H_3}C{H_2}C{H_3}\,(g)\, + \,{O_2}(g)\, \to \,C{O_2}\,(g)\, + \,{H_2}O(l)\, - - - - - - - - - - - - - - (1)$
Here atoms on the left hand side are, $(3C,\,8H\,and\,2O)$ and atoms on the right hand side are $(1C,\,2H\,and\,3O)$ . So as we see that number of atoms are less on right hand side so let’s multiply the carbon dioxide molecule by $3$ after that we get the equation as,
$C{H_3}C{H_2}C{H_3}\,(g)\, + \,{O_2}(g)\, \to \,3C{O_2}\,(g)\, + \,{H_2}O(l)\, - - - - - - - - - - - - - - (2)$
Now as we see that carbon get balanced having three carbons on both side, now we have to balanced hydrogens and oxygens by multiplying the water molecule by number $4$ and oxygen on left hand side by number $5$ .
$C{H_3}C{H_2}C{H_3}\,(g)\, + \,5{O_2}(g)\, \to \,3C{O_2}\,(g)\, + \,4{H_2}O(l)\, - - - - - - - - - - - - - - (3)$
The above equation is completely balanced, there are three carbon on both sides with $10\,oxygen$ on both sides. Also we have hydrogens as $8\,hydrogens$ . Thus complete balanced equation of combustion reaction between propane and oxygen is:
$C{H_3}C{H_2}C{H_3}\,(g)\, + \,5{O_2}(g)\, \to \,3C{O_2}\,(g)\, + \,4{H_2}O(l)$
Note:As we see that the number of atoms on both sides becomes equal. The same mechanism takes place when we have to balance the other hydrocarbons. We have two categories of hydrocarbons, saturated and unsaturated. In unsaturated one we have double or triple bonds while in saturated hydrocarbons only a single bond is present. Mainly we have to balance the carbons, hydrogens and oxygens in these types of reactions.
Complete step-by-step answer:We have propane which reacts with oxygen and performs a combustion reaction. The products of combustion reaction are carbon dioxide and water. So let’s start with counting the number of atoms left hand side and right hand side. We have equation as,
$C{H_3}C{H_2}C{H_3}\,(g)\, + \,{O_2}(g)\, \to \,C{O_2}\,(g)\, + \,{H_2}O(l)\, - - - - - - - - - - - - - - (1)$
Here atoms on the left hand side are, $(3C,\,8H\,and\,2O)$ and atoms on the right hand side are $(1C,\,2H\,and\,3O)$ . So as we see that number of atoms are less on right hand side so let’s multiply the carbon dioxide molecule by $3$ after that we get the equation as,
$C{H_3}C{H_2}C{H_3}\,(g)\, + \,{O_2}(g)\, \to \,3C{O_2}\,(g)\, + \,{H_2}O(l)\, - - - - - - - - - - - - - - (2)$
Now as we see that carbon get balanced having three carbons on both side, now we have to balanced hydrogens and oxygens by multiplying the water molecule by number $4$ and oxygen on left hand side by number $5$ .
$C{H_3}C{H_2}C{H_3}\,(g)\, + \,5{O_2}(g)\, \to \,3C{O_2}\,(g)\, + \,4{H_2}O(l)\, - - - - - - - - - - - - - - (3)$
The above equation is completely balanced, there are three carbon on both sides with $10\,oxygen$ on both sides. Also we have hydrogens as $8\,hydrogens$ . Thus complete balanced equation of combustion reaction between propane and oxygen is:
$C{H_3}C{H_2}C{H_3}\,(g)\, + \,5{O_2}(g)\, \to \,3C{O_2}\,(g)\, + \,4{H_2}O(l)$
Note:As we see that the number of atoms on both sides becomes equal. The same mechanism takes place when we have to balance the other hydrocarbons. We have two categories of hydrocarbons, saturated and unsaturated. In unsaturated one we have double or triple bonds while in saturated hydrocarbons only a single bond is present. Mainly we have to balance the carbons, hydrogens and oxygens in these types of reactions.
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