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What mass of Fe(OH)3 will be obtained when 100. mL of 0.240 M FeCl3 is mixed with 200. mL of 0.182 M NaOH?

Sagot :

Maacastrobridn

Answer:

648.68 mg

Explanation:

The reaction that takes place is:

  • FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl

First we calculate how many moles of each reactant were added, using the given volumes and concentrations:

  • FeCl₃ ⇒ 100 mL * 0.240 M = 24 mmol FeCl₃
  • NaOH ⇒ 100 mL * 0.182 M = 18.2 mmol NaOH

24 mmol of FeCl₃ would react completely with (24 * 3) 72 mmol of NaOH. There are not as many NaOH mmoles, so NaOH is the limiting reactant.

Now we calculate how many moles of Fe(OH)₃ are formed, using the moles of the limiting reactant:

  • 18.2 mmol NaOH * [tex]\frac{1mmolFe(OH)_3}{3mmolNaOH}[/tex] = 6.07 mmol Fe(OH)₃

Finally we convert 6.07 mmol Fe(OH)₃ to grams, using its molar mass:

  • 6.07 mmol Fe(OH)₃ * 106.867 mg/mmol = 648.68 mg

The 129.6 g mass of Fe(OH)₃ will be obtained.

How we calculate mass or weight from moles?

Weight or mass of any substance can be calculated from mole as:

n = W/M.

Given chemical reaction is:

FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl

According to the question:

Concentration of FeCl₃ = 0.240 M

Volume of FeCl₃ = 100 mL

Concentration of NaOH = 0.182 M

Volume of NaOH = 200 mL

Moles can be calculated by using the formula M = n/V mole/L 0r M = n×1000/V g/mL,

Moles of FeCl₃ = 0.240 × 100 = 0.024 mole

Moles of NaOH = 0.182 × 200 = 3.64 mole

From the stoichiometry of the reaction, it is clear that 3 mole of NaOH i.e. 109.2 moles of NaOH reacts with 1 mole of FeCl₃ to form product. So here NaOH is the limiting agent and responsible for the formation of product. Now we calculate how many moles of Fe(OH)₃ are formed, using the moles of the limiting reactant as follow:

Moles of Fe(OH)₃ = 1/3 × ( 3.64) = 1.21 mole

Finally we convert 12.1 mole Fe(OH)₃ to grams, using its molar mass:

W = 1.21 mole × 106.87 g/mole = 129.6 g

Hence, 129.6 gram of Fe(OH)₃ will be obtained.

To know more about moles, visit the below link:

https://brainly.com/question/14464650

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