Question

The following equation represents the combustion of

propane in excess oxygen.

C³h⁸ + 5O²----> 3CO² + 4H²O
What is the volume of carbon dioxide gas produced
when 48 cm of propane is completely burnt

a.28 cm³
b.48 cm³
c.96 cm³
d.144cm³​

Answer 1

3*0.35L = 1.05L

Step-by-step explanation:

For part A, we need the ratio of moles of propane to moles of oxygen to get the volume of oxygen required.

The equation tells us there are 5 moles of oxygen gas required to combust 1 mole of propane.

So the required volume of oxygen is 5*0.35L = 1.75L

The equation also tells us that 3 moles of carbon dioxide are produced from the combustion of one mole of propane.

Therefore the produced volume of carbon dioxide is 3*0.35L = 1.05L

Answer 2

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