3.04 liters of propane(C₃H₈) are needed to react with 15.2 liters of oxygen to produce water and carbon dioxide
Explanation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
The mole ratio between the propane and oxygen in the equation is 1 : 5 as can be seen in the equation.
If the amount of O₂ reacting is 15.2 L then the amount of propane required is;
1 : 5
X : 15.2
We cross multiply the ration
5x = 15.2 * 1
X = 15.2/5
X = 3.04
= 3.04 L
3.04 liters of propane(C₃H₈) are needed.
Balanced chemical equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
The mole ratio between the propane and oxygen in the equation is 1 : 5 as can be seen in the equation.
If the amount of O₂ reacting is 15.2 L then the amount of propane required is;
1 : 5
x : 15.2
For solving
5x = 15.2 * 1
x = 15.2/5
x = 3.04
Thus, the volume for propane needed is 3.04 L.
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