A gas has a volume of 6.35 L at 88.6 kPa. What will be the volume at standard pressure? Standard Pressure = 101.3 kPa
1 answer:
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32.8 g of Butane is required and 99.3 g of CO₂ is produced
<u>Explanation:</u>
The above mentioned reaction can be written as,
C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g) where ΔH (rxn)= -2658 kJ
It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.
That is
of butane reacted
Now this moles is converted into mass by multiplying it with its molar mass = 0.564 mol × 58.122 g / mol
= 32.8 g of butane.
Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol
= 99.3 g of CO₂
Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced