Answer:
Approximately 5.15 liters of H2O gas are produced when 7.25 liters of C3H8 are burned at STP
Step-by-step explanation:
The balanced chemical equation for the combustion of propane (C3H8) shows that for every 1 mole of C3H8 burned, 4 moles of H2O are produced. Using this information, we can set up the following conversion:
7.25 L C3H8 x 1 mol C3H8 / 22.4 L (at STP) x 4 mol H2O / 1 mol C3H8 x 22.4 L (at STP) / 1 mol H2O
Simplifying this expression, we get:
7.25 L C3H8 x 4 / 22.4 x 22.4 / 1 ≈ 5.15 L H2O gas
Therefore, approximately 5.15 liters of H2O gas are produced when 7.25 liters of C3H8 are burned at STP.