The volume of carbon dioxide produced if 8.45 g of propane is burned would be 12.9 L
Stoichiometric calculation
From the equation of the burning:
[tex]C_3H_8+ 5O_2 -- > 4H_2O+ 3CO_2[/tex]
The mole ratio of propane burned to carbon dioxide produced is 1:3.
Mole of 8.45 g propane = 8.45/44.1 = 0.19 moles
Equivalent mole of carbon dioxide produced = 0.19 x 3 = 0.57 moles
1 mole of any gas at STP = 22.4 L
0.57 moles of carbon dioxide at STP = 22.4 x 0.57 = 12.9 L
More on molar volume of gases at STP can be found here: https://brainly.com/question/1542685
