Question

Balance the following equation:
$Ca_{ 3 }(PO_{ 4 })_{ 2 }\quad +\quad SiO_{ 2 }\quad +\quad C\quad \rightarrow \quad CaSiO_{ 3 }\quad +\quad P_{ 4 }\quad +\quad CO$

Answer 1

Hint: This is an oxidation-reduction (redox) reaction. In order to balance a chemical equation, it is important to understand the law of conservation of mass. So, try to recall this law of conservation of mass.

Complete step by step answer:

Here are the rules for balancing a chemical equation [oxidation number method] -

Write an unbalanced equation first to start our approach:
$Ca_{ 3 }(PO_{ 4 })_{ 2 }\quad +\quad SiO_{ 2 }\quad +\quad C\quad \rightarrow \quad CaSiO_{ 3 }\quad +\quad P_{ 4 }\quad +\quad CO$

Assign oxidation number for each atom and Identify redox reactions and write separately as half reactions.
L.H.S :
            $Ca_{ 3 }(PO_{ 4 })_{ 2 }$ = Ca(+2) , P(+5) , O(-2)
            $SiO_{ 2 }$ = Si(+4) , O(-2)
            C = 0
            R.H.S :
            $CaSiO_{ 3 }$ = Ca(+2) , Si(+4) , O(-2)
            $P_{ 4 }$ = P(0)
            CO = C(+2) , O(-2)

Oxidation number of P, +5 (in $Ca_{ 3 }(PO_{ 4 })_{ 2 }$) changes to 0 (in $P_{ 4 }$). Hence it is a reduction half reaction.
$P^{ V }+5e-\quad \rightarrow \quad P^{ 0 }$
Oxidation number of C, 0 (in C) changes to +2 (in $CO_2$). Hence it is an oxidation half reaction.
$C^{ 0 }\quad -\quad 2e-\quad \rightarrow \quad C^{ II }$
Here $Ca_{ 3 }(PO_{ 4 })_{ 2 }$ is an oxidizing agent, C is a reducing agent.

Make electron gain equivalent to electron loss in the half reactions by multiplying the above reaction with 2 and 5 respectively.
$2P^{ V }+10e-\quad \rightarrow \quad 2P^{ 0 }$ (reduction half reaction)
$5C^{ 0 }\quad -\quad 10e-\quad \rightarrow \quad 5C^{ II }$ (oxidation half reaction)
The number we multiplied each reaction with, will now be coefficient for each molecule. Now the reaction would look like this

$2Ca_{ 3 }(PO_{ 4 })_{ 2 }\quad +\quad SiO_{ 2 }\quad +\quad 5C\quad \rightarrow \quad CaSiO_{ 3 }\quad +\quad P_{ 4 }\quad +\quad 5CO$

Balance all other atoms except H and O in the reaction.
$2Ca_{ 3 }(PO_{ 4 })_{ 2 }\quad +\quad 6SiO_{ 2 }\quad +\quad 5C\quad \rightarrow \quad 6CaSiO_{ 3 }\quad +\quad P_{ 4 }\quad +\quad 5CO$

Balance the hydrogen by H+ or OH− and balance the oxygen atoms by adding $H_2O$.
We will get the final equation as:
$2Ca_{ 3 }(PO_{ 4 })_{ 2 }\quad +\quad 6SiO_{ 2 }\quad +\quad 10C\quad \rightarrow \quad 6CaSiO_{ 3 }\quad +\quad P_{ 4 }\quad +\quad 10CO$

And this is the correct answer for the question.

Note: For any chemical equation (in a closed system) the mass of the reactants must be equal to the mass of the products. In order to make sure that this is the case, the number of atoms of each element in the reactants must be equal to the number of atoms of those same elements in the products.