To find the percent yield of water, we need to follow a series of steps:
1. Write the balanced chemical equation for the reaction.
2. Calculate the theoretical yield of water based on the limiting reagent.
3. Calculate the percent yield using the actual yield and the theoretical yield.
### Step 1: Write the Balanced Chemical Equation
The balanced chemical equation for the combustion of ethane is:
\[ \text{2 C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 4 \text{CO}_2 + 6 \text{H}_2\text{O} \]
### Step 2: Find the Limiting Reagent
To determine the limiting reagent, we need to calculate the moles of each reactant.
#### Moles of Ethane (C\(_2\)H\(_6\)):
\[
\text{Molar mass of C}_2\text{H}_6 = 2(12.01) + 6(1.01) = 30.08 \,\text{g/mol}
\]
\[
\text{Moles of C}_2\text{H}_6 = \frac{4.5 \, \text{g}}{30.08 \, \text{g/mol}} = 0.1495 \, \text{mol}
\]
#### Moles of Oxygen (O\(_2\)):
\[
\text{Molar mass of O}_2 = 2(16.00) = 32.00 \,\text{g/mol}
\]
\[
\text{Moles of O}_2 = \frac{5.5 \, \text{g}}{32.00 \, \text{g/mol}} = 0.1719 \, \text{mol}
\]
The balanced reaction tells us that 2 moles of C\(_2\)H\(_6\) require 7 moles of O\(_2\):
\[
\text{Moles of O}_2 \, \text{needed for given C}_2\text{H}_6 = 0.1495 \, \text{mol} \times \frac{7}{2} = 0.523 \, \text{mol}
\]
Since 0.523 moles of O\(_2\) are required but only 0.1719 moles are available, O\(_2\) is the limiting reagent.
### Step 3: Calculate Theoretical Yield of Water
Using the stoichiometry from the balanced equation, 7 moles of O\(_2\) produce 6 moles of H\(_2\)O:
\[
\text{Moles of H}_2\text{O} \, \text{produced} = 0.1719 \, \text{mol O}_2 \times \frac{6}{7} = 0.147 \, \text{mol H}_2\text{O}
\]
Convert moles of H\(_2\)O to grams:
\[
\text{Molar mass of H}_2\text{O} = 2(1.01) + 16.00 = 18.02 \,\text{g/mol}
\]
\[
\text{Theoretical yield} = 0.147 \, \text{mol H}_2\text{O} \times 18.02 \,\text{g/mol} = 2.65 \, \text{g}
\]
### Step 4: Calculate Percent Yield
Percent yield formula:
\[
\text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100\%
\]
\[
\text{Percent yield} = \left( \frac{1.86 \, \text{g}}{2.65 \, \text{g}} \right) \times 100\% = 70.2\%
\]
So, the percent yield of water is \(70.2\%\).
Gaseous ethane (CH3CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 1.86 g of water is produced from the reaction of 4.5 g of ethane and 5.5 g of oxygen gas, calculate the percent yield of water.
Be sure your answer has the correct number of significant digits in it.
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