In lab you reacted 60 grams of propane (C3H8) with 13 grams of oxygen gas (O2). Your reaction made 63 grams of water (H2O). The reaction is shown below:

How much carbon dioxide (CO2) is released into the air?

C3H8 + 5O2 > 3CO2 + 4H2O

10 g

73 g

120 g

1368

120g

Explanation:

Based on the reaction, when 1 mole of propane and 5 moles of oxygen react, 3 moles of CO2 and 4 moles of H2O are produced. The ratio of production is 3 moles of CO2 per 4 moles of H2O.

Thus, we need to convert mass of water to moles using its molar mass:

Moles H2O (Molar mass: 18g/mol):

63g H2O * (1mol / 18g) = 3.5 moles H2O

Converting to moles of CO2:

3.5 moles H2O * (3 moles CO2 / 4 moles H2O) = 2.625 moles CO2

Mass CO2 (Molar mass: 44g/mol):

2.625 moles CO2 * (44g / mol) = 115.5g of CO2 are released

the answer is barium and copper bromide