How many moles of iron(III) hydroxide precipitate will form when 2.7 moles of aqueous sodium hydroxide reacts completely with excess iron(III) nitrate solution according to the following reaction? (record your answer to 1 decimal place)

0.9 mole of Fe(OH)3.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

Fe(NO3)3 + 3NaOH —> Fe(OH)3 + 3NaNO3

Now, we can determine the moles of iron (III) hydroxide formed from the reaction as follow:

From the balanced equation above,

3 moles of NaOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 2.7 moles of NaOH will react to produce = 2.7/3 = 0.9 mole of Fe(OH)3.

Therefore, 0.9 mole of Fe(OH)3 is produced from the reaction.

yes it is balanced