Answer :
Answer:
32.1 g
Explanation:
Step 1: Write the balanced combustion reaction
C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O
Step 2: Calculate the moles corresponding to 97.4 g of CO₂
The molar mass of CO₂ is 44.01 g/mol.
97.4 g × 1 mol/44.01 g = 2.21 mol
Step 3: Calculate the moles of butane that produced 2.21 moles of carbon dioxide
The molar ratio of C₄H₁₀ to CO₂ is 1:4. The moles of C₄H₁₀ required are 1/4 × 2.21 mol = 0.553 mol
Step 4: Calculate the mass corresponding to 0.553 moles of C₄H₁₀
The molar mass of C₄H₁₀ is 58.12 g/mol.
0.553 mol × 58.12 g/mol = 32.1 g
Final answer:
To produce 97.4 g of CO2 from butane, approximately 32.1 g of butane would be required when going through the stages of molar calculations.
Explanation:
First, it is important to know that the balanced chemical equation for the combustion of Butane (C4H10) is: C4H10 + (13 / 2)O2 -> 4CO2 + 5H2O. This tells us that for every mole of butane combusted, 4 moles of carbon dioxide are produced.
Given that the molar mass of carbon dioxide (CO2) is 44.01 g/mol, we can calculate the number of moles in 97.4 g of CO2: 97.4 g / 44.01 g/mol = 2.213 moles of CO2.
Since 4 moles of CO2 are formed from 1 mole of butane, the moles of butane needed are: 2.213 moles CO2 * 1 mol C4H10 / 4 mol CO2 = 0.5533 moles of C4H10.
Finally, given that the molar mass of butane (C4H10) is approximately 58.12 g/mol, the mass of butane needed is: 0.5533 moles C4H10 * 58.12 g/mol C4H10 = 32.1 g of butane, presented to three significant figures.
Learn more about Calculating Mass in Chemistry here:
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