Question

li₃po₄ → li + p + o₂
cs₂ + f₂ → cf₄ + s
fe + cl₂ → fecl₃

Explanation:

Step1: Balance the first equation

For the reaction $\mathrm{Li_3PO_4}\to\mathrm{Li}+\mathrm{P}+\mathrm{O_2}$, we need to balance the number of each atom on both sides. There are 3 Li - atoms in $\mathrm{Li_3PO_4}$, so we put a 3 in front of Li on the product - side. There is 1 P - atom and 4 O - atoms in $\mathrm{Li_3PO_4}$. To balance the O - atoms, since we have $\mathrm{O_2}$ on the product - side, we need 2 moles of $\mathrm{O_2}$ to get 4 O - atoms. The balanced equation is $\mathrm{Li_3PO_4}\to3\mathrm{Li}+\mathrm{P}+2\mathrm{O_2}$.

Step2: Balance the second equation

For the reaction $\mathrm{CS_2}+\mathrm{F_2}\to\mathrm{CF_4}+2\mathrm{S}$, we first balance the S - atoms. There are 2 S - atoms in $\mathrm{CS_2}$, so we put a 2 in front of S on the product - side. Then, to balance the F - atoms, since there are 4 F - atoms in $\mathrm{CF_4}$, we need 2 moles of $\mathrm{F_2}$ on the reactant - side. The balanced equation is $\mathrm{CS_2}+2\mathrm{F_2}\to\mathrm{CF_4}+2\mathrm{S}$.

Step3: Balance the third equation

For the reaction $\mathrm{Fe}+\mathrm{Cl_2}\to\mathrm{FeCl_3}$, first, we balance the Cl - atoms. There are 2 Cl - atoms in $\mathrm{Cl_2}$ and 3 Cl - atoms in $\mathrm{FeCl_3}$. The least - common multiple of 2 and 3 is 6. So we put a 3 in front of $\mathrm{Cl_2}$ and a 2 in front of $\mathrm{FeCl_3}$ to get 6 Cl - atoms on both sides. Then, to balance the Fe - atoms, we put a 2 in front of Fe. The balanced equation is $2\mathrm{Fe}+3\mathrm{Cl_2}\to2\mathrm{FeCl_3}$.

Answer:

Balanced equations:
$\mathrm{Li_3PO_4}\to3\mathrm{Li}+\mathrm{P}+2\mathrm{O_2}$
$\mathrm{CS_2}+2\mathrm{F_2}\to\mathrm{CF_4}+2\mathrm{S}$
$2\mathrm{Fe}+3\mathrm{Cl_2}\to2\mathrm{FeCl_3}$