Question

How many milliliters of 0.200 M NH₄OH are needed to react with 12.0 mL of 0.550 M FeCl₃?

FeCl₃ + 3NH₄OH → Fe(OH)₃ + 3NH₄Cl

Answer 1

Answer) 99 millilitres are needed.

Explanation:

1 mL = 0.001 L

Volume = 12.0 mL ×(0.001 L/1 mL) =0.012 L

Calculate the number of moles of FeCl_​​​3 using the equation as follows:

Number of moles = molarity × volume in litres

Number of moles = 0.550 M × 0.012 L = 0.0066 mol FeCl_​​​3

According to the balanced chemical equation, 1 mol FeCl_​​​3 reacts with 3 mol NH_{​​​​​4}​OH.

1 mol FeCl_​​​3 = 3 mol NH_{​​​​​4}​OH

0.0066 mol FeCl_​​​3 = 0.0066 × 3 mol NH_{​​​​​4}​OH

0.0066 mol FeCl_​​​3 = 0.0198 mol NH_{​​​​​4}​OH

Calculate the volume of NH_{​​​​4}OH in litres as follows:

Volume = number of moles/molarity

Volume = 0.0198 mol / 0.200 M = 0.099 L

Volume = 0.099 L ×(1 mL/0.001 L) =99 mL