3KCI + Fe(NO3)3 —> 3KNO3 + FeCI3. Question: If 10.0 grams of Fe(NO3)3 is used, how many grams of of KCI are required?

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03-11-2022
Chemistry

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3KCI + Fe(NO3)3 —> 3KNO3 + FeCI3. Question: If 10.0 grams of Fe(NO3)3 is used, how many grams of of KCI are required?

Sagot :

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AzirF141584 AzirF141584 · Answer 1 · 2022-11-03T21:02:57-04:00

Answer:

9.24 g of KCl.

Explanation:

What is given?

Mass of Fe(NO3)3 = 10.0 g.

Molar mass of Fe(NO3)3 = 242 g/mol.

Mlar mass of KcCl = 74.5 g/mol. (you can calculate the molar massesof a compound using the periodc table)

Step-by-step solution:

Let's covert 110.0 g of Fe(NO3)3 to moles using is molar mass, like this:

[tex]1\text{0.0 g Fe\lparen NO}_3)_3\cdot\frac{1\text{ mol Fe\lparen NO}_3)_3}{242\text{ g Fe\lparen NO}_3)_3}=0.0413\text{ moles Fe\lparen NO}_3)_3.[/tex]

You can see in the chemical equation that 3 moles of KCl are required to react with 1 mol of Fe(NO3)3, so by doing a rule o three to find how many moles of KClowe require to react with 0.0413 moles Fe(NO3)3, the calculation will look this:

[tex]0.0413\text{ moles Fe\lparen NO}_3)_3\cdot\frac{3\text{ moles KCl}}{1\text{ mol Fe\lparen NO}_3)_3}=0.124\text{ moles KCl.}[/tex]

And the final step is to convert 0.124 moles of KCl to grams using its given molar mass, likethis:

[tex]0.124\text{ moles KCl}\cdot\frac{74.5\text{ g KCl}}{1\text{ mol KCl}}=9.238\text{ g KCl}\approx9.24\text{ g KCl.}[/tex]

We will require 9.24 g of KCl to react with 10.0 g of Fe(NO3)3.

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