which is the correctly balanced equation for the reaction of rust, fe₂o₃, and hydrochloric acid, hcl?\no…

Answer:

C. $\mathrm{Fe_2O_3}+6\mathrm{HCl}\rightarrow 2\mathrm{FeCl_3}+3\mathrm{H_2O}$

Explanation:

Step1: Balance iron atoms

On the left - hand side, there are 2 Fe atoms in $\mathrm{Fe_2O_3}$. In $\mathrm{FeCl_3}$, to balance Fe, we put a coefficient of 2 in front of $\mathrm{FeCl_3}$: $\mathrm{Fe_2O_3}+\mathrm{HCl}\rightarrow 2\mathrm{FeCl_3}+\mathrm{H_2O}$.

Step2: Balance chlorine atoms

In $2\mathrm{FeCl_3}$, there are 6 Cl atoms. So we put a coefficient of 6 in front of $\mathrm{HCl}$: $\mathrm{Fe_2O_3}+6\mathrm{HCl}\rightarrow 2\mathrm{FeCl_3}+\mathrm{H_2O}$.

Step3: Balance hydrogen and oxygen atoms

On the left - hand side, we have 6 H atoms from 6 $\mathrm{HCl}$ and 3 O atoms from $\mathrm{Fe_2O_3}$. On the right - hand side, in $2\mathrm{FeCl_3}$, there are no H and O atoms. In $\mathrm{H_2O}$, to balance H and O, we put a coefficient of 3 in front of $\mathrm{H_2O}$. So the balanced equation is $\mathrm{Fe_2O_3}+6\mathrm{HCl}\rightarrow 2\mathrm{FeCl_3}+3\mathrm{H_2O}$.