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2C₂H₆ (g) + 7O₂ (g) --> 4CO₂ (g) + 6H₂O (g)
You mix 16.5 grams of ethane (C₂H₆) with 17 grams of oxygen gas (O₂).
a) Which gas is the limiting reactant reactant?
b) How many grams of water should be produced?
c) How many grams of the excess reactant is left over?
d) If only 6.6 grams of water are produced, what is the percent yield?

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D) If only 6.6 grams of water are produced, what is the percent yield?

Write out the balanced equation:

2C2H6 + 7O2 --> 4CO2 + 6H2O

Then, determine the limiting reagents:

16.8 grams of C2H6/30.06 g/mol of C2H6 = 0.46 mol

45.8 grams of O2/32 g/mol of O2 = 1.4 mol

0.46 mol of C2H6/2 = 0.23

1.4 mol of O2/7 = 0.20

- Dioxygen is the limiting reagent.  

Find the theoretical molar yield of water:

There is a 7 mol of O2 to 6 mol of H2O ratio (according to the balanced equation).

We can setup a proportion;

7/6 = 0.23/x

0.196 = x

So, 0.196 mol of water will be produced.

How many grams of water will be produced in the reaction?

0.196 mol of H2O x 18 g/mol of H2O = 3.53 grams of H2O

znk

Answers:

(a) Oâ‚‚ 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  

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

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 Oâ‚‚ = 0.531 mol Oâ‚‚

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

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 Hâ‚‚O = 1.646 mol Hâ‚‚O

From Oâ‚‚:

The molar ratio of Hâ‚‚O:Oâ‚‚ is 6 mol Hâ‚‚O:7 mol Oâ‚‚.

Moles of H₂O = 0.531 × 6/7

Moles of Hâ‚‚O = 0.455 mol Hâ‚‚O

The limiting reactant is Oâ‚‚ because it gives the smaller amount of Hâ‚‚O.

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

Step 4. Calculate the theoretical yield of Hâ‚‚O that you can obtain from Oâ‚‚.

Theoretical yield of H₂O = 0.455 × 18.02/1

Theoretical yield of Hâ‚‚O = 8.2 g Hâ‚‚O

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

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

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

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

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

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 %

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