Question

magnesium burns in the presence of oxygen to form magnesium oxide. Write chemical equation and balance it​

Answer 1

Answer:

When magnesium ribbon burns in presence of oxygen. We could observe a dazzling white flame and a white substance named magnesium oxide.

2Mg+O2  ⟶2MgO

Answer 2

Question:

Magnesium burns in the presence of Oxygen to form magnesium oxide. Write chemical equation and balance it.

Answer:

Magnesium burns in the presence of Oxygen to form magnesium oxide. This reaction can be written in words as:

$\rm{Magnesium+Oxygen\rightarrow Magnesium \: Oxide}$

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Now,

Symbol of Magnesium is $\rm{Mg}$

Formula of Oxygen is $\rm{{O}_{2}}$

Formula of Magnesium Oxide is $\rm{MgO}$

So, putting the symbols and Formulae of all substance in the above word equation. We get,

$\rm{Mg+{O}_{2}\rightarrow MgO}$

Now, we have to balance the chemical equation.

Let us count the number of atoms of all the elements in the reactants and product separately.

$\begin{array}{| c | c | c |}\hline{} & \sf{In \: reactants} & \sf{In \: products} \\ \hline{}\\ \sf{Number \: \: of \: \: Magnesium} \: \: atoms & 1 & 1 \\ \\ \hline{} \\ \sf{Number \: \: of \: \: Oxygen} \: atoms & 2 & 1 \\ \\ \hline\end{array}$

We find that the Reactant contains one Magnesium atom and product also contains one Magnesium atom. Also, the Reactant contains two Oxygen atoms and product contains one oxygen atom. The above reaction contains unequal number of Oxygen atoms in reactant and product, so it is an unbalanced equation.

Now, to have two Oxygen atoms on right side of the equation , we multiply Mgo by 2 and write it as 2MgO. so that:

$\rm{Mg+{O}_{2}\rightarrow 2MgO}$

Let us count the number of atoms on both side of the equation again.

$\begin{array}{| c | c | c |}\hline{} & \sf{In \: reactants} & \sf{In \: products} \\ \hline{} \\ \sf{Number \: \: of \: \: Magnesium} \: atoms & 1 & 2 \\ \\ \hline{} \\ \sf{Number \: \: of \: \: Oxygen} \: \: atoms & 2 & 2 \\ \\ \hline\end{array}$

Though the number of Oxygen atoms has become equal on both side but the number of Magnesium atoms has now become unequal. There are two magnesium atom on right side of equation But, one atom on it's left side. To have two magnesium atom on left side , we multiply Mg by 2 and write 2Mg, so that:

$\rm{2Mg+{O}_{2}\rightarrow 2MgO}$

Let us count the number of atoms on both the sides again::

$\begin{array}{| c | c | c |} \hline{} & \sf{In \: reactants} & \sf{In \: products} \\ \hline{} \\ \sf{Number \: \: of \: \: Magnesium} \: \: atoms & 2 & 2 \\ \\ \hline{} \\ \sf{Number \: \: of \: \: Oxygen} \: \: atoms & 2 & 2 \\ \\ \hline\end{array}$

This chemical equation now contains an equal number of atoms of Magnesium and oxygen on both the side. So, this is a balanced equation.

Balanced Equation:

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \dashrightarrow\underline{\color{purple}{\rm{2Mg+{O}_{2}\longrightarrow 2MgO}}}$