Question

60 points please help me i would appreciate it!

Answer 1

`\huge\boxed{\sf 39\ L}`

Step-by-step explanation:

In the reaction:

for 1 mol of propane, 3 moles of CO₂ are produced.

Volume of propane = 13 L

At STP,

1 mol of propane = 22.4 L/dm³ molar value

So,

Moles in 13 L of propane:

= volume / molar volume

= 13 / 22.4

= 0.58 mol

Since,

1 mol of propane = 3 moles of CO₂

0.58 mol of propane = 3 × 0.58 mol of CO₂

0.58 mol of propane = 1.74 mol of CO₂

Converting 1.74 mol of CO₂ into L:

[Multiply it by 22.4 L/mol]

= 1.74 mol × 22.4 L/mol

= 39 L

`\rule[225]{225}{2}`