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Math  /  Numbers & Operations

QuestionGaseous ethane (CH3CH3)\left(\mathrm{CH}_{3} \mathrm{CH}_{3}\right) will react with gaseous oxygen (O2)\left(\mathrm{O}_{2}\right) to produce gaseous carbon dioxide (CO2)\left(\mathrm{CO}_{2}\right) and gaseous water (H2O)\left(\mathrm{H}_{2} \mathrm{O}\right). Suppose 21. g of ethane is mixed with 124. g of oxygen. Calculate the minimum mass of ethane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits. □\square g

Studdy Solution

STEP 1

1. The chemical reaction is a combustion reaction of ethane with oxygen.
2. The balanced chemical equation for the reaction is: $ 2 \mathrm{C}_2\mathrm{H}_6 + 7 \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2 + 6 \mathrm{H}_2\mathrm{O} \]
3. The molar mass of ethane (\(\mathrm{C}_2\mathrm{H}_6\)) is approximately \(30.07 \, \text{g/mol}\).
4. The molar mass of oxygen (\(\mathrm{O}_2\)) is approximately \(32.00 \, \text{g/mol}\).
5. We need to determine the limiting reactant to find the minimum mass of ethane left over.

STEP 2

1. Calculate moles of ethane and oxygen.
2. Determine the limiting reactant.
3. Calculate the mass of ethane left over.

STEP 3

Calculate the moles of ethane (C2H6\mathrm{C}_2\mathrm{H}_6).
Moles of C2H6=21.0 g30.07 g/mol≈0.698 mol\text{Moles of } \mathrm{C}_2\mathrm{H}_6 = \frac{21.0 \, \text{g}}{30.07 \, \text{g/mol}} \approx 0.698 \, \text{mol}

STEP 4

Calculate the moles of oxygen (O2\mathrm{O}_2).
Moles of O2=124.0 g32.00 g/mol=3.875 mol\text{Moles of } \mathrm{O}_2 = \frac{124.0 \, \text{g}}{32.00 \, \text{g/mol}} = 3.875 \, \text{mol}

STEP 5

Determine the limiting reactant using the stoichiometry of the balanced equation: 2 mol C2H6:7 mol O22 \, \text{mol} \, \mathrm{C}_2\mathrm{H}_6 : 7 \, \text{mol} \, \mathrm{O}_2
Calculate the required moles of O2\mathrm{O}_2 for the available C2H6\mathrm{C}_2\mathrm{H}_6:
Required moles of O2=0.698 mol×72=2.443 mol\text{Required moles of } \mathrm{O}_2 = 0.698 \, \text{mol} \times \frac{7}{2} = 2.443 \, \text{mol}
Since 2.443 mol<3.875 mol2.443 \, \text{mol} < 3.875 \, \text{mol}, C2H6\mathrm{C}_2\mathrm{H}_6 is the limiting reactant.

STEP 6

Since C2H6\mathrm{C}_2\mathrm{H}_6 is the limiting reactant, no ethane is left over after the reaction.
Mass of ethane left over is:
0 g\boxed{0 \, \text{g}}

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