Chemistry, 13.01.2021 09:40 drubio102004

How much excess reactant is left over when 17.0 g of potassium hydroxide (KOH) reacts with 20.0 g of iron (III) nitrate (Fe(NO3)2)?

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Answer from: Anahartie

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

Answer from: Quest

what is the average time it took the car to travel 0.25 meters?

⇒ 2.23 seconds

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How much excess reactant is left over when 17.0 g of potassium hydroxide (KOH) reacts with 20.0 g o...