Studdy AI Logo
Math  /  Algebra

Question1. Write a balanced chemical equation, including physical state symbols, for the combustion of liquid nonane into gaseous carbon dioxide and gaseous water. C9H20(l)+14O2(g)→9CO2(g)+10H2O(g)\mathrm{C}_{9} \mathrm{H}_{20}(l)+14 \mathrm{O}_{2}(g) \rightarrow 9 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)
2. Suppose 0.340 kg of nonane are burned in air at a pressure of exactly 1 atm and a temperature of 15.0∘C15.0^{\circ} \mathrm{C}. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits. □\square L

Studdy Solution

STEP 1

1. The chemical reaction is given as: $ \mathrm{C}_{9} \mathrm{H}_{20}(l) + 14 \mathrm{O}_{2}(g) \rightarrow 9 \mathrm{CO}_{2}(g) + 10 \mathrm{H}_{2} \mathrm{O}(g) \]
2. We assume complete combustion of nonane.
3. The ideal gas law is applicable for calculating the volume of carbon dioxide.
4. The molar mass of nonane (\(\mathrm{C}_{9} \mathrm{H}_{20}\)) is approximately \(128.26 \, \text{g/mol}\).
5. The temperature is converted to Kelvin for gas law calculations.

STEP 2

1. Calculate the moles of nonane.
2. Use stoichiometry to find the moles of carbon dioxide produced.
3. Apply the ideal gas law to find the volume of carbon dioxide.

STEP 3

Calculate the moles of nonane using its molar mass. Given that the mass of nonane is 0.340 kg0.340 \, \text{kg}, convert this to grams:
0.340 kg=340 g 0.340 \, \text{kg} = 340 \, \text{g}
Next, calculate the moles of nonane:
Moles of nonane=340 g128.26 g/mol \text{Moles of nonane} = \frac{340 \, \text{g}}{128.26 \, \text{g/mol}}
Moles of nonane≈2.65 mol \text{Moles of nonane} \approx 2.65 \, \text{mol}

STEP 4

Use the balanced chemical equation to find the moles of carbon dioxide produced. The equation shows that 1 mole of nonane produces 9 moles of carbon dioxide:
Moles of CO2=2.65 mol×9 \text{Moles of } \mathrm{CO}_{2} = 2.65 \, \text{mol} \times 9
Moles of CO2≈23.85 mol \text{Moles of } \mathrm{CO}_{2} \approx 23.85 \, \text{mol}

STEP 5

Use the ideal gas law to calculate the volume of carbon dioxide. The ideal gas law is given by:
PV=nRT PV = nRT
Where: - P=1 atm P = 1 \, \text{atm} - V=? L V = ? \, \text{L} - n=23.85 mol n = 23.85 \, \text{mol} - R=0.0821 L atm/mol K R = 0.0821 \, \text{L atm/mol K} - T=15.0∘C=288.15 K T = 15.0^{\circ} \mathrm{C} = 288.15 \, \text{K}
Rearrange the ideal gas law to solve for V V :
V=nRTP V = \frac{nRT}{P}
V=23.85 mol×0.0821 L atm/mol K×288.15 K1 atm V = \frac{23.85 \, \text{mol} \times 0.0821 \, \text{L atm/mol K} \times 288.15 \, \text{K}}{1 \, \text{atm}}
V≈563.5 L V \approx 563.5 \, \text{L}
Round to 3 significant digits:
V≈564 L V \approx 564 \, \text{L}
The volume of carbon dioxide gas produced is:
564 L \boxed{564} \, \text{L}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord