Final answer:
To calculate the number of liters of oxygen required to form 10.5 g of H2O at STP, we need to convert grams to moles of water and then use a mole ratio to find the moles of oxygen. Finally, we can use the ideal gas law to calculate the volume of oxygen, which is 6.97 L.
Step-by-step explanation:
To calculate the number of liters of oxygen required to form 10.5 g of H2O at STP (Standard Temperature and Pressure), we need to use the molar mass of water and convert grams to moles.
From the balanced chemical equation, we can see that 2 moles of H2O is formed for every 1 mole of O2.
Therefore, we can set up a ratio:
2 mol H2O : 1 mol O2
Now, let's calculate the moles of H2O:
Moles of H2O = mass of H2O / molar mass of H2O
Moles of H2O = 10.5 g / 18.015 g/mol
Next, we can use the mole ratio to find the moles of O2:
Moles of O2 = (moles of H2O) / 2
Finally, we can use the ideal gas law to calculate the volume of O2:
Volume of O2 = (moles of O2) x (22.4 L/mol)
Plugging in the values, we get:
Volume of O2 = (10.5 g / 18.015 g/mol) / 2 x 22.4 L/mol
Volume of O2 = 6.97 L