Question

the combustion of propane may be described by the chemical equation c3h8(g) 5o2(g)⟶3co2(g) 4h2o(g) how many grams of o2(g) are needed to completely burn 80.5 g c3h8(g)?

Answer 1

To completely burn 80.5 g of C3H8, you would need 292 g of O2.

Step-by-step explanation:

To determine how many grams of O2(g) are needed to completely burn 80.5 g C3H8(g), we need to use the stoichiometry of the balanced chemical equation. The balanced equation is:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

According to the stoichiometry, 1 mole of C3H8 reacts with 5 moles of O2. To find the moles of O2 needed to burn 80.5 g of C3H8, we can use the molar mass of C3H8 to convert grams to moles:

Molar mass of C3H8 = 3(12.01 g/mol) + 8(1.01 g/mol) = 44.11 g/mol

Moles of C3H8 = 80.5 g C3H8 × (1 mol C3H8/44.11 g C3H8) = 1.826 mol C3H8

Using the stoichiometry of the balanced equation, we can determine the moles of O2 needed:

1.826 mol C3H8 × (5 mol O2/1 mol C3H8) = 9.13 mol O2

To convert moles to grams, we can use the molar mass of O2:

Molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol

Grams of O2 needed = 9.13 mol O2 × (32.00 g O2/1 mol O2) = 292 g O2