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Answer :

Maacastrobrgo

Answer:

520 mg Al(OH)₃

Explanation:

The reaction that takes place is:

  • Al(NO₃)₂ + 3KOH → Al(OH)₃ + 3KNO₃

Now we calculate how many moles of each reagent were added, using their concentrations and volumes:

  • Al(NO₃)₂ ⇒ 50.0 mL * 0.200 M = 10 mmol Al(NO₃)₂
  • KOH ⇒ 200.0 mL * 0.100 M = 20 mmol KOH

10 mmol of Al(NO₃)₂ would react completely with (10*3) 30 mmol of KOH, there are not as many KOH mmol, so KOH is the limiting reactant.

We calculate how many moles of Al(OH)₃ are produced, using the limiting reactant:

  • 20 mmol KOH * [tex]\frac{1mmolAl(OH)_3}{3mmolKOH}[/tex] = 6.67 mmol Al(OH)₃

Finally we convert mmol of Al(OH)₃ to mg, using its molar mass:

  • 6.67 mmol Al(OH)₃ * 78 mg/mmol = 520 mg Al(OH)₃

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