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elena-s [515]
3 years ago
11

98 POINTS!!! Must SHOW YOUR WORK!

Chemistry
2 answers:
abruzzese [7]3 years ago
6 0

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

<em>- Dioxygen is the limiting reagent.  </em>

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).

<em>We can setup a proportion;</em>

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?

<em><u>0.196 mol of H2O x 18 g/mol of H2O = 3.53 grams of H2O </u></em>

Inga [223]3 years ago
4 0

Answers:

(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 <em>limiting reactant problem</em>.  

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

Step 1. <em>Gather all the information in one place</em> 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 <em>moles of each reactant</em>  

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

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

Step 3. <em>Identify the limiting reactant </em>

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

<em>From C₂H₆ :</em>

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

<em>From O₂: </em>

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.

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

Step 4. Calculate the <em>theoretical yield of H₂O</em> that you can obtain from O₂.

Theoretical yield of H₂O = 0.455 × 18.02/1

Theoretical yield of H₂O = 8.2 g H₂O

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

Step 5. Calculate the <em>moles of C₂H₆ consumed</em>

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 <em>mass of C₂H₆ consumed. </em>

Mass of C₂H₆ = 0.130 × 30.07

Mass of C₂H₆ = 3.91 g C₂H₆

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

Step 7. Calculate the <em>mass of unreacted C₂H₆</em>

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 <em>percent yield </em>

% Yield = actual yield/theoretical yield × 100 %

Actual yield = 6.6 g

% yield = 6.6/8.2 × 100

% yield = 80 %

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Explanation:

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      Where  k_B is the Boltzmann constant with a value of 1.38*10^{-23} kg \cdot m^2 /s^2 \cdot ^o K

                    T is the room temperature

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Generally number of  atoms in mass of an element can be obtained using the mathematical operation

                      n = \frac{m}{M}  * N_A

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             m = 1 kg = 1 kg *  \frac{10000 g}{1kg }  = 1000g

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            C= 392 J/kg\cdot ^o K

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