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
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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
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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 %