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How would you balance the following equation? $Ca\, + \,HN{O_3}\, \to \,Ca{(N{O_3})_2}\, + \,{N_2}{O_3}\, + \,{H_2}O$

Last updated date: 20th Jun 2024
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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, atoms are hydrogen oxygen calcium nitrogen we have found that equation in which each atom is balanced at last.

Complete step-by-step answer:In the above equation we have the reaction in between calcium and nitric acid which forms calcium nitrate and dinitrogen trioxide with water. Now the first step while solving the equation for balancing we have to count the number of atoms on both sides of the equation. We have $(1\,Ca,\,1H,\,1N\,and\,3O)$ on left hand side of equation while on right hand side we have, $(1\,Ca,\,2H,\,4N\,and\,10O)$ as you are seeing that there are lesser number of atoms on left hand side let’s try to change the stoichiometry of reactants.
$Ca\, + \,HN{O_3}\, \to \,Ca{(N{O_3})_2}\, + \,{N_2}{O_3}\, + \,{H_2}O - - - - - - - - - - - - (1)$
Let’s multiply the left side it means the reactant $HN{O_{3\,}}\,with\,6$ and $Ca\,with\,2$ we get the equation as
 $2Ca\, + \,6HN{O_3}\, \to \,Ca{(N{O_3})_2}\, + \,{N_2}{O_3}\, + \,{H_2}O\, - - - - - - - - - - (2)$ . Now to make the calcium atom equal on both sides we have to multiply the right hand side calcium nitrate $Ca{(N{O_3})_2}\,with\,2$ . $2Ca\, + \,6HN{O_3}\, \to \,2Ca{(N{O_3})_2}\, + \,{N_2}{O_3}\, + \,{H_2}O\, - - - - - - - - - - (3)$
See the equation and count the number of atoms of each type, there are two calcium each on both sides, six nitrogen on both sides, but now oxygen and hydrogen are unbalanced. So, for making the number of hydrogens and oxygens equal we multiply the water molecule on right hand side with $3$ We get the balanced equation as number $(4)$ as
$2Ca\, + \,6HN{O_3}\, \to \,2Ca{(N{O_3})_2}\, + \,{N_2}{O_3}\, + \,3{H_2}O\, - - - - - - - - - - (4)$ Here, calcium becomes equal, hydrogens are six, nitrogen’s are also six in number, oxygen are eighteen on both side.

Note:The balancing of chemical equations is very important because by this process we get to know the moles of reactant used and products formed. Here, we get an idea that $2\,moles\,of\,Ca$ reacts with $6\,moles\,of\,nitric\,acid$ and gives $2\,mole\,of\,calcium\,nitrate\,and\,3\,mole\,of\,water$ . There are other methods also available for balancing like oxidation reduction method and electron method.