Question

If 252 grams of iron are reacted with 321 grams of chlorine gas, what is the mass of the excess reactant leftover after the reaction has reached completion?
2Fe(s)+3Cl2(g)⟶2FeCl3(s)

If 152 grams of ethane (C2H6) are reacted with 231 grams of oxygen gas, what is the excess reactant?
2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(g)

Answer 1

Answer:

Iron is in excess.

1) The mass of the iron remaining = 83.38 grams

2) Ethane is in excess. There will remain 90.06 grams ethane

Explanation:

Step 1: Data given

Mass of iron = 252 grams

Mass of Cl2 = 321 grams

Molar mass of Fe = 55.845

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Fe(s)+3Cl2(g)⟶2FeCl3(s)

Step 3: Calculate moles

Moles = mass / molar mass

Moles Fe = 252.0 grams / 55.845 g/mol = 4.512 moles

Moles Cl2 = 321.0 grams / 70.90 g/mol = 4.528 moles

Step 4: Calculate the limiting reactant

For 2 moles Fe we need 3 moles Cl2 to produce 2 moles Fecl3

Cl2 is the limiting reactant. It will completely be consumed (4.528 moles).

Fe is in excess. There will 4.528 * 2/3 = 3.019 moles be consumed

There will remain 4.512 - 3.019 = 1.493 moles of Fe

The mass of the iron remaining = 1.493 * 55.845 g/mol =83.38 grams

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If 152 grams of ethane (C2H6) are reacted with 231 grams of oxygen gas, what is the excess reactant?

Step 1: Data given

Mass of ethane = 152.0 grams

mass of O2 =231.0 grams

Molar mass of ethane = 30.07 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

2C2H6(g) + 7O2(g) ⟶ 4CO2(g) +  6H2O(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles ethane = 152.0 grams / 30.07 g/mol = 5.055 moles

Moles O2 = 231.0 grams / 32.0 g/mol = 7.22 moles

Step 4: Calculate limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed (7.22 moles).

Ethane is in excess. There will react 7.22 * 2/7 = 2.06 moles

There will remain 5.055 - 2.06 = 2.995 moles ethane

2.995 moles ethane = 2.995 * 30.07 g/mol = 90.06 grams ethane

Answer 2

Answer:

Explanation:

(a) O₂ is the limiting reactant.

(b) The theoretical yield of water is 8.2 g.

(c) The mass of unreacted C₂H₆ is 12.6 g.

(d) The percent yield of water is 80 %.

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.  

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

Step 1. Gather all the information in one place with molar masses above the formulas and masses below them.  

M_r:          30.07   32.00                  18.02

                2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O

Mass/g:      16.5       17  

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Step 2. Calculate the moles of each reactant  

Moles of C₂H₆ = 16.5 × 1/30.07

Moles of C₂H₆ = 0.5487 mol C₂H₆

Moles of O₂ = 17 × 1/32.00

Moles of O₂ = 0.531 mol O₂

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Step 3. Identify the limiting reactant

Calculate the moles of H₂O we can obtain from each reactant.  

From C₂H₆ :

The molar ratio of H₂O:C₂H₆ is 6 mol H₂O:2 mol C₂H₆

Moles of H₂O = 0.5487 × 6/2

Moles of H₂O = 1.646 mol H₂O

From O₂:

The molar ratio of H₂O:O₂ is 6 mol H₂O:7 mol O₂.

Moles of H₂O = 0.531 × 6/7

Moles of H₂O = 0.455 mol H₂O

The limiting reactant is O₂ because it gives the smaller amount of H₂O.

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Step 4. Calculate the theoretical yield of H₂O that you can obtain from O₂.

Theoretical yield of H₂O = 0.455 × 18.02/1

Theoretical yield of H₂O = 8.2 g H₂O

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Step 5. Calculate the moles of C₂H₆ consumed

The molar ratio of C₂H₆:O₂ is 2 mol C₂H₆:7 mol O₂.

Moles of C₂H₆ = 0.455 × 2/7

Moles of C₂H₆ = 0.130 mol C₂H₆

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Step 6. Calculate the mass of C₂H₆ consumed.

Mass of C₂H₆ = 0.130 × 30.07

Mass of C₂H₆ = 3.91 g C₂H₆

===============

Step 7. Calculate the mass of unreacted C₂H₆

Starting mass = 16.5 g

Mass consumed = 3.91 g

Mass unreacted = 16.5 – 3.91

Mass unreacted = 12.6 g

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Step 8. Calculate the percent yield

% Yield = actual yield/theoretical yield × 100 %

Actual yield = 6.6 g

% yield = 6.6/8.2 × 100

% yield = 80 %