Respuesta :
Answer: The mass of iron (III) chloride produced is 14.81 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For iron(III) oxide:
Given mass of iron(III) oxide = 10.0 g
Molar mass of iron(III) oxide = 159.7 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of iron(III) oxide}=\frac{10.0g}{159.7g/mol}=0.0626mol[/tex]
- For hydrochloric acid:
Given mass of hydrochloric acid = 10.0 g
Molar mass of hydrochloric acid = 36.5 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of hydrochloric acid}=\frac{10.0g}{36.5g/mol}=0.274mol[/tex]
The chemical equation for the reaction of iron (III) oxide and hydrochloric acid follows:
[tex]Fe_2O_3+6HCl\rightarrow 2FeCl_3+3H_2O[/tex]
By Stoichiometry of the reaction:
6 moles of hydrochloric acid reacts with 1 mole of iron (III) oxide
So, 0.274 moles of hydrochloric acid will react with = [tex]\frac{1}{6}\times 0.274=0.0456mol[/tex] of iron (III) oxide
As, given amount of iron (III) oxide is more than the required amount. So, it is considered as an excess reagent.
Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
6 moles of hydrochloric acid produces 2 moles of iron (III) chloride
So, 0.274 moles of hydrochloric acid will produce = [tex]\frac{2}{6}\times 0.274=0.0913moles[/tex] of iron (III) chloride
Now, calculating the mass of iron (III) chloride from equation 1, we get:
Molar mass of iron (III) chloride = 162.2 g/mol
Moles of iron (III) chloride = 0.0913 moles
Putting values in equation 1, we get:
[tex]0.0913mol=\frac{\text{Mass of iron (III) chloride}}{162.2g/mol}\\\\\text{Mass of iron (III) chloride}=(0.0913mol\times 162.2g/mol)=14.81g[/tex]
Hence, the mass of iron (III) chloride produced is 14.81 grams
