Question

\\_\\_\\_\\_ c₃h₈ + \\_\\_\\_\\_ o₂ \\_\\_\\_\\_
\\_\\_\\_\\_ fecl₃ + \\_\\_\\_\\_ naoh → \\_\\_\\_\\_ fe(oh)₃ + \\_\\_\\_\\_ nacl

Explanation:

Step1: Balance the iron atoms

For the reaction $\mathrm{FeCl}_{3}+\mathrm{NaOH}
ightarrow\mathrm{Fe(OH)}_{3}+\mathrm{NaCl}$, there is 1 iron atom on the left - hand side and 1 on the right - hand side, so the iron is already balanced in terms of the number of atoms.

Step2: Balance the chlorine atoms

There are 3 chlorine atoms in $\mathrm{FeCl}_{3}$ on the left - hand side. In $\mathrm{NaCl}$ on the right - hand side, to balance the chlorine atoms, we need 3 moles of $\mathrm{NaCl}$. So the equation becomes $\mathrm{FeCl}_{3}+\mathrm{NaOH}
ightarrow\mathrm{Fe(OH)}_{3}+3\mathrm{NaCl}$.

Step3: Balance the sodium and oxygen atoms

There are 3 sodium atoms in 3 moles of $\mathrm{NaCl}$ on the right - hand side. So we need 3 moles of $\mathrm{NaOH}$ on the left - hand side to balance the sodium atoms. Also, when we have 3 moles of $\mathrm{NaOH}$, the oxygen and hydrogen atoms are also balanced. The balanced chemical equation is $\mathrm{FeCl}_{3}+3\mathrm{NaOH}
ightarrow\mathrm{Fe(OH)}_{3}+3\mathrm{NaCl}$.

For the reaction $\mathrm{C}_{3}\mathrm{H}_{8}+\mathrm{O}_{2}
ightarrow$ (products not fully shown in the problem, assuming complete combustion to $\mathrm{CO}_{2}$ and $\mathrm{H}_{2}\mathrm{O}$):

Step1: Balance the carbon atoms

In $\mathrm{C}_{3}\mathrm{H}_{8}$, there are 3 carbon atoms. So we need 3 moles of $\mathrm{CO}_{2}$ on the right - hand side: $\mathrm{C}_{3}\mathrm{H}_{8}+\mathrm{O}_{2}
ightarrow3\mathrm{CO}_{2}+\mathrm{H}_{2}\mathrm{O}$.

Step2: Balance the hydrogen atoms

In $\mathrm{C}_{3}\mathrm{H}_{8}$, there are 8 hydrogen atoms. In $\mathrm{H}_{2}\mathrm{O}$, to balance the hydrogen atoms, we need 4 moles of $\mathrm{H}_{2}\mathrm{O}$ since $2\times4 = 8$. So the equation becomes $\mathrm{C}_{3}\mathrm{H}_{8}+\mathrm{O}_{2}
ightarrow3\mathrm{CO}_{2}+4\mathrm{H}_{2}\mathrm{O}$.

Step3: Balance the oxygen atoms

On the right - hand side, there are $(3\times2)+(4\times1)=6 + 4=10$ oxygen atoms. So we need 5 moles of $\mathrm{O}_{2}$ on the left - hand side. The balanced chemical equation is $\mathrm{C}_{3}\mathrm{H}_{8}+5\mathrm{O}_{2}
ightarrow3\mathrm{CO}_{2}+4\mathrm{H}_{2}\mathrm{O}$.

Answer:

For $\mathrm{C}_{3}\mathrm{H}_{8}+\mathrm{O}_{2}
ightarrow$: 1, 5, $3\mathrm{CO}_{2}+4\mathrm{H}_{2}\mathrm{O}$
For $\mathrm{FeCl}_{3}+\mathrm{NaOH}
ightarrow$: 1, 3, 1, 3