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How many moles of Iron III hydroxide are formed when 0.385 moles of potassium hydroxide react with Iron III chloride?

Sagot :

Naǫidn
[tex]3 \ KOH + FeCl_3 \to Fe(OH)_3 + 3 \ KCl[/tex]

3 moles of KOH - 1 mole of Fe(OH)₃
0.385 moles of KOH - x moles of Fe(OH)₃
cross multiply:
[tex]3x=1 \times 0.385 \\ 3x=0.385 \ \ \ |\div 3 \\ x \approx 0.128[/tex]

0.128 moles of iron(III) hydroxide are formed.
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