Answer :
[tex]98.56 dm^3[/tex] of oxygen at STP would be required to react completely with 38.8g of propane.
Given that :
molar mass of propane = 44 g/mol
mass of propane = 38.8 g
∴ Moles present in 38.8 g of propane = [tex]\frac{38.8}{44}[/tex] = 0.88 mole
applying rule of balanced equations
1 mole of propane = 5 moles of oxygen
0.88 mole of propane = 5 * 0.88 = 4.4 moles of oxygen
Note : volume of 1 mole of oxygen at STP = [tex]22.4 dm^3[/tex]
∴Total volume of oxygen required at STP = 22.4 * 4.4 = [tex]98.56 dm^3[/tex]
Hence we can conclude that the volume of oxygen at STP required to react completely [tex]98.56 dm^3[/tex]
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Final answer:
To combust 38.8g of propane completely, 98.56 dm³ of oxygen at STP are required, calculated using stoichiometry and the molar volume of gases.
Explanation:
To find out how many dm³ of oxygen at STP are required to react completely with 38.8 grams of propane, we need to follow these steps:
Calculate the moles of propane (C3H8) using its molar mass (C3H8 = 44.1 g/mol).
Apply the stoichiometry of the chemical reaction, using the balanced equation for the combustion of propane: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l). This equation tells us that 1 mole of propane reacts with 5 moles of oxygen.
Use the molar volume concept to convert moles of oxygen to volume at STP, where 1 mole of any gas occupies 22.4 dm³.
Let’s do this calculation:
1. Moles of propane: 38.8 g / 44.1 g/mol = 0.88 moles of C3H8.
2. Moles of oxygen needed: 0.88 moles of C3H8 × 5 moles O2/mole C3H8 = 4.4 moles O2.
3. Volume of oxygen at STP: 4.4 moles O2 × 22.4 dm³/mole = 98.56 dm³.
Therefore, 98.56 dm³ of oxygen at STP would be required to react completely with 38.8 g of propane.
