Answer :
If 3.45 g of sodium hydroxide is being reacted with 5.23 g of iron (iii) chloride, grams of iron (iii) hydroxide produced is 3.44 g.
The reaction is given as :
FeCl₃ + 3NaOH ---> Fe(OH)₃ + 3NaCl
mass of NaOH = 3.45 g
mass of FeCl₃ = 5.23 g
moles of NaOH = mass / molar mass
= 3.45 / 40
= 0.086 mol
moles of FeCl₃ = 5.23 / 162
= 0.0322 mol
FeCl₃ is the limiting reactant and depends on the production of the Fe(OH)₃ .
1 mole of FeCl₃ = 1 mole of Fe(OH)₃
0.0322 mol of FeCl₃ = 0.0322 mol of Fe(OH)₃
mass of Fe(OH)₃ = moles × molar mass
= 0.0322 × 107
= 3.44 g
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