Question

how many liters of oxygen at STP are required to burn completely 2.2 g of propane, C3H8?​

Answer 1

Answer:-

5.6 L of oxygen at STP are required to burn completely 2.2g of propane.

Explanation:-

The balanced chemical equation for the reaction is :-

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

_______________________________

• Molar mass of Carbon (C) = 12 g/mol

• Molar mass of Hydrogen (H) = 1 g/mol

Hence, molar mass of Propane [C₃H₈] :-

= 12×3 + 1×8

= 36 + 8

= 44 g/mol

Now, from the balnced equation :-

⇒ 44g of C₃H₈ requires 5 moles of O₂

We know that, one mole of any gas at STP occupies a volume of 22.4 L.

∴ 5 mole of O₂ will occupy a volume of :-

= 5×22.4

= 112 L

∴ 44g C₃H₈ requires 112 L O₂

∴ 2.2g of propane will require :-

= 2.2×112/44

= 0.05×112

= 5.6 L O₂

Answer 2

Explanation:

Answer:-✍️✍️

5.6 L of oxygen at STP are required to burn completely 2.2g of propane.

Explanation:-

The balanced chemical equation for the reaction is :-

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

_______________________________

• Molar mass of Carbon (C) = 12 g/mol

• Molar mass of Hydrogen (H) = 1 g/mol

Hence, molar mass of Propane [C₃H₈] :-

= 12×3 + 1×8

= 36 + 8

= 44 g/mol

Now, from the balnced equation :-

⇒ 44g of C₃H₈ requires 5 moles of O₂

We know that, one mole of any gas at STP occupies a volume of 22.4 L.

∴ 5 mole of O₂ will occupy a volume of :-

= 5×22.4

= 112 L

∴ 44g C₃H₈ requires 112 L O₂

∴ 2.2g of propane will require :-

= 2.2×112/44

= 0.05×112

= 5.6 L O₂

hope this helps you ✨

thank you ⭐