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kipiarov [429]
3 years ago
9

When iron(III) oxide reacts with hydrochloric acid, iron(III) chloride and water are formed. How many grams of iron(III) chlorid

e are formed from 10.0 g of iron(III) oxide and 10.0 g of hydrochloric acid
Chemistry
1 answer:
Aleksandr [31]3 years ago
7 0

<u>Answer:</u> The mass of iron (III) chloride produced is 14.81 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For iron(III) oxide:</u>

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:

\text{Moles of iron(III) oxide}=\frac{10.0g}{159.7g/mol}=0.0626mol

  • <u>For hydrochloric acid:</u>

Given mass of hydrochloric acid = 10.0 g

Molar mass of hydrochloric acid = 36.5 g/mol

Putting values in equation 1, we get:

\text{Moles of hydrochloric acid}=\frac{10.0g}{36.5g/mol}=0.274mol

The chemical equation for the reaction of iron (III) oxide and hydrochloric acid follows:

Fe_2O_3+6HCl\rightarrow 2FeCl_3+3H_2O

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 = \frac{1}{6}\times 0.274=0.0456mol 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 = \frac{2}{6}\times 0.274=0.0913moles 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:

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

Hence, the mass of iron (III) chloride produced is 14.81 grams

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