Question

Write the balanced equation for the following chemical reactions,
A. $F{e_2}{O_3} + \,Al\, \to Fe + A{l_2}{O_3}$
B. $Mg + HCl \to MgC{l_2} + \,{H_2}$

Answer 1

For balancing the reaction you firstly have to see that the number of atoms of each type are the same on both sides or not. If the number is the same then it is already balanced but in case where the number of atoms are not the same, you have to make them equal by multiplying the stoichiometry of reaction and products.

Complete step-by-step answer: First we have the reaction between ferric oxide and aluminum and they react and form iron and aluminum oxide. Here the number of atoms on both sides are different that is the number of atom son left hand side are $(2Fe,\,1Al,\,3O)$ and on the right hand side we have $(1Fe,\,2Al,\,3O)$ so now we need to balance the number. If we multiply free aluminium by the number $2$ and similarly iron by $2$ we get the equation as below and it is balanced.
$F{e_2}{O_3} + \,2Al\, \to 2Fe + A{l_2}{O_3}$
Next we have a reaction between magnesium and hydrochloric acid which in reaction form magnesium chloride and release hydrogen gas. Here also the number is different for atoms on both sides, $(1Mg,\,1Cl,\,1H)$ on the left hand side and $(1Mg,\,2Cl,\,2H)$ on the right hand side. If we multiply the reactants side where there is hydrochloric acid by number $2$ then we get the balanced chemical equation.
$Mg + 2HCl \to MgC{l_2} + \,{H_2}$

Note: For making a chemical reaction balanced there are many methods like oxidation reduction method and half-reaction method by which we can make it balanced. At the end we get the number of atoms on both sides equal. As shown above we can multiply that side where the numbers are less, this will balance them.