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How many moles of iron (III) hydroxide are produced when 5.0 moles of sodium hydroxide reacts with excess iron (III) chloride? FeCl3 + 3NaOH —> Fe(OH)3 + 3NaCl

User Kbgn
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Answer:

1.67 moles of iron (III) hydroxide are produced when 5.0 moles of sodium hydroxide reacts with excess iron (III) chloride.

Step-by-step explanation:

The balanced reaction is:

FeCl₃ + 3 NaOH ⇒ Fe(OH)₃+ 3 NaCl

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagents and products participate in the reaction:

  • FeCl₃: 1 mole
  • NaOH: 3 moles
  • Fe(OH)₃: 1 mole
  • NaCl: 3 moles

Then you can apply the following rule of three: if 3 moles of NaOH form 1 mole of Fe(OH)₃, 5 moles of NaOH how many moles of Fe(OH)₃ do they form?


moles of Fe(OH)_(3) =(5 moles of NaOH*1 mole of Fe(OH)_(3) )/(3 moles of NaOH)

moles of Fe(OH)₃= 1.67

1.67 moles of iron (III) hydroxide are produced when 5.0 moles of sodium hydroxide reacts with excess iron (III) chloride.

User Alireza Alallah
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