Part C: Questions

1. Mercury reacts with oxygen according to the following reaction: 2Hg+O?₂ →2HgO
2. What would be the limiting reactant if you had 6 mol of oxygen and 12 mol of mercury??
3. We have looked at mol ratios, which can be converted to molecules using Avagadro's number or masses of reactants using molar masses. If you had 500 g of mercury (Hg) and 100 g of oxygen gas (O? ), which would be the limiting reactant? (Mercury's atomic mass is 200.6 u and the atomic 2? mass for a single oxygen atom is 15.999 u) ?SHOW YOUR CALCULATIONS

3. If you reacted 1 mol of aluminum (Al) with 1 mol of iodine gas (I? ), what amount of product could be 2? formed? (aluminum's atomic mass is 26.98 u and the atomic mass for a single iodine atom is 126.90u).?SHOWYOURCALCULATIONS

2Al+3I?₂→2Al(I)?₃

4. A bicycle company has 35 wheels and 20 frames to build bicycles. Which is the "limiting reactant" which will determine how many bikes they can assemble? ?

 

Question 1

Answer:

a) Both Hg and O₂ are limiting reactants.

b) Hg is the limiting reactant.

 

Question 2

Answer:

Amount of product formed = 272 grams of Al(I)₃ or 0.667 moles of Al(I)₃

 

Question 3

Answer:

limiting reactant = wheels

Step-by-step explanation

For the given three problems, the main topic we need to know is the limiting reactant.

 

In simple terms, the limiting reactant is the chemical species CONSUMED FIRST in a given reaction. The amount of limiting reactant dictates how much product will be produced. Once the limiting reactant is completely consumed, the reaction will stop while the remaining reactants after the reaction will be in excess. They are called excess reactants.

 

Now, to determine the limiting reactant, we can use the easier approach which is the method of cases.

Here, we will assume 2 cases. For each case, one of the reactant will be assumed as the limiting reactant.

Now, the case which produces the LEAST products will be the limiting reactant.

 

This method comes from the fact that the limiting reactant "limits" the product that will be produced. Thus, the limiting reactant will make the reaction produce less products. So, we will just assume that each reactant is a limiting reactant and determine the "true limiting reactant" from the amount of product it will produce.

 

To better understand, let us perform them to the given problems.

 

Question 1:

We are given the chemical equation,

2Hg+O?₂ →2HgO

a) We are asked what would be the limiting reactant if we have 6 moles of O₂ (oxygen) and 12 moles of Hg (mercury).

b) Given 500 g of mercury (Hg) and 100 g of oxygen (O₂), determine the limiting reactant.

 

From the method we discussed earlier, first, we will have 2 cases.

 

CASE 1: Hg is the limiting reactant

CASE 2: O₂ is the limiting reactant

 

Let us go to the first case.

CASE 1: Hg is the limiting reactant

[Using the given moles of Hg and the balanced reaction, we will determine the moles of HgO that can be produced.]

 

moles of HgO produced = (moles of Hg) x mole ratio

moles of HgO = 12 moles Hg (2 moles HgO / 2 moles Hg)

[Note: The mole ratio, (2 moles HgO / 2 moles Hg) came from the balanced reaction.

Simplifying,

moles of HgO = 12 moles

 

CASE 2: O₂ is the limiting reactant

 

moles of HgO produced = (moles of O₂) x mole ratio

moles of HgO = 6 moles O₂ (2 moles HgO / 1 mole O₂)

[Note: (2 moles HgO / 2 moles Hg)

Simplifying,

moles of HgO = 12 moles

 

Now, since both cases produced both 12 moles of HgO, it means that BOTH REACTANTS (Hg and O₂) are LIMITING REACTANTS.

 

 

Let us go to the letter b.

Given 500 g of mercury (Hg) and 100 g of oxygen (O₂), determine the limiting reactant.

 

Again, we will assume 2 cases.

 

CASE 1: Hg is the limiting reactant

CASE 2: O₂ is the limiting reactant

 

Let us go to the first case.

CASE 1: Hg is the limiting reactant

 

This time, we will use the molar mass of each reactant to convert their given masses to moles and determine the moles of product it will produce.

 

moles of HgO produced = (mass of Hg / molar mass of Hg) x mole ratio

moles of HgO produced = (500 g Hg / 200.6 g/mol) (2 moles HgO / 2 moles Hg)

Simplifying,

moles of HgO = 2.5 moles

 

CASE 2: O₂ is the limiting reactant

 

moles of HgO produced = (mass of O₂ / molar mass of O₂) x mole ratio

moles of HgO = (100 g O₂ / 32.00 g/mol O₂) x (2 moles HgO / 1 mole O₂)

Simplifying,

moles of HgO = 6.25 moles

 

Since CASE 1 produced less moles of HgO, the limiting reactant will be Hg (mercury).

 

Now, let us proceed to the second question.

Question 2:

We are given the reaction,

2Al+3I?₂→2Al(I)₃

Then, we are asked the amount of product formed when 1 moles of Al reacts with 1 mole of I₂ according to the given balanced chemical equation.

 

Again, we will have 2 cases.

 

CASE 1: Al is the limiting reactant

CASE 2: I₂ is the limiting reactant

 

Let us go to the first case.

CASE 1: Al is the limiting reactant

 

moles of Al(I)₃ produced = (moles of Al) x mole ratio

moles of Al(I)₃ = 1 mole Al (2 mole Al(I)₃ / 2 mole Al)

Simplifying,

moles of Al(I)₃= 1 mole

 

CASE 2: I₂ is the limiting reactant

 

moles of Al(I)₃ produced = (moles of I₂) x mole ratio

moles of Al(I)₃ produced = 1 mole I₂ (2 moles Al(I)₃ / 3 mole I₂)

Simplifying,

moles of Al(I)₃= 0.6667 moles

 

Since CASE 2 produced less moles of Al(I)₃, the limiting reactant will be I₂ (iodine) and the moles of product formed is 0.667 moles.

 

Lastly, we will convert 0.6667 moles Al(I)₃ to mass.

mass of Al(I)₃ = moles of Al(I)₃ x molar mass of Al(I)₃

mass of Al(I)₃ = (0.6667 moles) x (26.98 g/mol + 3 x 126.90 g/mol)

mass of Al(I)₃ = 272 grams (final answer)

 

Let us now proceed to Question 3.

We are given 35 wheels and 20 frames. Then, we are asked which will be the "limiting reactant".

 

Now, we know that to assemble a bike, we need 2 wheels and 1 frame.

If we transform this to an equation,

2 wheels + 1 frams ---> 1 bike

 

Now, just like last time, we will have 2 cases.

 

CASE 1: Wheel is the limiting reactant

CASE 2: Frame is the limiting reactant

 

Let us proceed to CASE 1.

CASE 1: Wheel is the limiting reactant

number of bikes produced = number of wheels x mole ratio

number of bikes produced = 35 wheels (1 bike / 2 wheels)

Simplifying,

number of bikes produced = 17.5 bikes

 

Let us proceed to CASE 2.

CASE 2: Frame is the limiting reactant

number of bikes produced = number of frames x mole ratio

number of bikes produced = 20 frames (1 bike / 1 frame)

Simplifying,

number of bikes produced = 20 bikes

 

Since CASE 2 produced less number of bikes, the "limiting reactant" will be the wheels.