Question

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)?

Answer 1

3.10 g

Step-by-step explanation:

The formula for this reaction is:

3KOH + Fe(NO3)3 --> 3KNO3 + Fe(OH)3

1) Convert grams to moles

17.0 g KOH ÷ 56.11g = 0.303 mol

20.0 g Fe(NO3)3 ÷ 241.86g = 0.0827 mol

2) Determine limiting and excess reactants by dividing the number of moles by the molar coefficient

0.303÷3=0.101 mol KOH

0.0827÷1=0.0827 mol Fe(NO3)

Since Fe(NO3)3 has the lowest amount, it will be the LR.

3) Find how many grams of KOH (the excess reactant) is used during the reaction

0.0827 mol Fe(NO3)3 × 3 mol KOH/1 mol Fe(NO3)3 × 56.11 g KOH/mol = 13.9 g KOH

4) Find how many grams are left over

The original amount of KOH we had was 17.0 g so subtract this by the amount used

17.0g-13.9g= 3.10 g KOH

Answer 2

4.56 g of KOH

Step-by-step 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