Answer:
4.56 g of KOH
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2KOH + Fe(NO₃)₂ —> Fe(OH)₂ + 2KNO₃
Next, we shall determine the masses of KOH and Fe(NO₃)₂ that reacted from the balanced equation. This is can be obtained as:
Molar mass of KOH = 39 + 16 + 1 = 56 g/mol
Mass of KOH from the balanced equation = 2 × 56 = 112 g
Molar mass of Fe(NO₃)₂ = 56 + 2[14 + (16×3)]
= 56 + 2[14 + 48)]
= 56 + 2[62]
= 56 + 124
= 180 g/mol
Mass of Fe(NO₃)₂ from the balanced equation = 1 × 180 = 180 g
SUMMARY:
From the balanced equation above,
112 g of KOH reacted with 180 g of Fe(NO₃)₂
Next, we shall determine the limiting reactant and the excess reactant. This can be obtained as follow:
From the balanced equation above,
112 g of KOH reacted with 180 g of Fe(NO₃)₂.
Therefore, 17 g of KOH will react with = (17 × 180)/112 = 27.32 g of Fe(NO₃)₂
From the calculations made above, we can see that it will take a higher mass (i.e 27.32 g) of Fe(NO₃)₂ than what was given (i.e 20 g) to react completely with 17 g of KOH.
Therefore, Fe(NO₃)₂ is the limiting reactant and KOH is the excess reactant.
Next, we shall determine the mass of the excess reactant that reacted. This can be obtained as follow:
From the balanced equation above,
112 g of KOH reacted with 180 g of Fe(NO₃)₂.
Therefore Xg of KOH will react with 20 g of Fe(NO₃)₂ i.e
Xg of KOH = (112 × 20)/180
Xg of KOH = 12.44 g
Thus, 12.44 g of KOH reacted.
Finally, we shall determine the leftover mass of the excess reactant.
The excess reactant is KOH. The leftover mass can be obtained as follow:
Mass of KOH given = 17 g
Mass of KOH that reacted = 12.44 g
Mass of KOH leftover =?
Mass of KOH leftover = (Mass of KOH given) – (Mass of KOH that reacted)
Mass of KOH leftover = 17 – 12.44
Mass of KOH leftover = 4.56 g
Thus, the excess reactant (i.e KOH) that is left over is 4.56 g
