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Ket [755]
2 years ago
10

According to the following reaction, how many moles of Fe(OH)2 can form from 175.0 mL of 0.227 M LiOH solution? Assume that ther

e is excess FeCl2. FeCl2(aq) + 2 LiOH(aq) → Fe(OH)2(s) + 2 LiCl(aq)
Chemistry
1 answer:
IgorC [24]2 years ago
3 0

Answer:

0.020 moles of Fe(OH)_{2} can be formed

Explanation:

1. First determine the number of moles of LiOH.

Molarity is given by the following expression:

M=\frac{molesofsolute}{Litersofsolution}

Solving for moles of solute:

moles of solute = M * Liters of solution

Converting 175.0mL to L:

175.0mL*\frac{1L}{1000mL}=0.175L

Replacing values:

moles of solute = 0.227M*0.175L

moles of solute = 0.040

Therefore there are 0.040 moles of LiOH

2. Then write the balanced chemical reaction and use the stoichiometry of the reaction to calculate the number of moles of Fe(OH)_{2} produced:

FeCl_{2}(aq)+2LiOH(aq)=Fe(OH)_{2}(s)+2LiCl(aq)

As the problem says that there are excess of FeCl_{2}, the limiting reagent is the LiOH.

0.040molesLiOH*\frac{1molFe(OH)_{2}}{2molesLiOH}=0.020molesFe(OH)_{2} can be formed

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<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

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We are given:

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We know that:

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In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, \frac{12}{44}\times 9.06=2.47g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, \frac{2}{18}\times 5.58=0.62g of hydrogen will be contained.

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To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = \frac{0.206}{0.206}=1

For Hydrogen  = \frac{0.62}{0.206}=3

For Oxygen  = \frac{0.206}{0.206}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

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For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

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Putting values in above equation, we get:

n=\frac{124g/mol}{31g/mol}=4

Multiplying this valency by the subscript of every element of empirical formula, we get:

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