Answer:
0.35 g.
Explanation:
We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.
This is illustrated below:
Molarity of Fe(NO3)3 = 0.250 M
Volume = 35 mL = 35/1000 = 0.035 L
Mole of Fe(NO3)3 =?
Molarity = mole /Volume
0.250 = mole of Fe(NO3)3 / 0.035
Cross multiply
Mole of Fe(NO3)3 = 0.25 x 0.035
Mole of Fe(NO3)3 = 8.75×10¯³ mole
Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M
KOH solution. This is illustrated below:
Molarity of KOH = 0.180 M
Volume = 55 mL = 55/1000 = 0.055 L
Mole of KOH =.?
Molarity = mole /Volume
0.180 = mole of KOH /0.055
Cross multiply
Mole of KOH = 0.180 x 0.055
Mole of KOH = 9.9×10¯³ mole.
Next, we shall write the balanced equation for the reaction. This is given below:
3KOH + Fe(NO3)3 —> Fe(OH)3 + 3KNO3
From the balanced equation above,
3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
3 moles of KOH reacted with 1 mole of Fe(NO3)3.
Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.
From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.
Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.
Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.
In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.
The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:
From the balanced equation above,
3 moles of KOH reacted to produce 1 mole of Fe(OH)3.
Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.
Finally, we shall convert 3.3×10¯³ mole of Fe(OH)3 to grams. This can be obtained as follow:
Molar mass of Fe(OH)3 = 56 + 3(16 + 1) = 56 + 3(17) = 107 g/mol
Mole of Fe(OH)3 = 3.3×10¯³ mole
Mass of Fe(OH)3 =?
Mole = mass /Molar mass
3.3×10¯³ = Mass of Fe(OH)3 / 107
Cross multiply
Mass of Fe(OH)3 = 3.3×10¯³ x 107
Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.
Therefore, 0.35 g of Fe(OH)3 was produced from the reaction.
is 44 g/mol and molar mass of
is 18 g/mol
