How many liters of water are formed from 2.9 kg of octane at STP? Be sure your answer has the correct number of significant figures.
Note: Reference the SI prefixes table for additional information.
To determine the volume of water formed from 2.9 kg of octane at STP, we first need to calculate the number of moles of octane using its molar mass.
The molar mass of octane (C8H18) is 114.22 g/mol.
2.9 kg = 2900 g
Number of moles of octane = 2900 g / 114.22 g/mol = 25.39 mol
According to the balanced chemical equation for the combustion of octane:
2C8H18 + 25O2 -> 16CO2 + 18H2O
The mole ratio between octane and water is 2:18.
Therefore, the number of moles of water formed is:
25.39 mol octane * (18 mol water / 2 mol octane) = 227.91 mol water
Now, we can use the ideal gas law to calculate the volume of water at STP (Standard Temperature and Pressure: 0 degrees Celsius and 1 atm pressure).
1 mol of any gas at STP occupies 22.4 liters.
Volume of water formed = 227.91 mol * 22.4 L/mol = 5100.56 L
Therefore, 2.9 kg of octane at STP will form 5100.56 liters of water.
The answer has 5 significant figures, which is the same as the precision of the given mass of octane (2.9 kg).