Final answer:
To find the number of moles of water vapor formed when 10 liters of butane gas is burned in oxygen at STP, we need to determine the balanced equation for the combustion of butane gas and use the ideal gas law equation. The balanced equation is C4H10 + 13/2 O2 -> 4 CO2 + 5 H2O. From this equation, we find that 4 moles of water vapor are formed for every mole of butane gas burned. Using the ideal gas law equation, we convert the given volume of butane gas to moles and then use the mole ratio to calculate the number of moles of water vapor formed.
Step-by-step explanation:
First, we need to determine the balanced equation for the combustion of butane gas:
C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
According to the balanced equation, 4 moles of water vapor are formed for every mole of butane gas burned. So, to calculate the number of moles of water vapor formed, we need to find the number of moles of butane gas present. We can use the ideal gas law equation to convert the given volume of butane gas to moles:
10 L × (1 mol / 22.414 L) = 0.4464 moles of butane gas
Since the mole ratio between butane gas and water vapor is 1:4, the number of moles of water vapor formed is:
0.4464 moles × (4 moles / 1 mole) = 1.7856 moles of water vapor