Final answer:
To find the volume of O2 gas required for the reaction with 70.25 grams of KH, we use stoichiometry and the ideal gas law. We first calculate the moles of KH, then find the corresponding moles of O2 needed, and finally calculate the volume using the ideal gas law at the given conditions of temperature and pressure.
Step-by-step explanation:
To determine how many liters of O2 gas at 25°C and 1.00 atm pressure are needed to react completely with 70.25 grams of potassium hydride (KH), we must use the given chemical equation:
2 KH (s) + O2 (g) → H2O (l) + K2O (s)
First, we calculate the moles of KH using its molar mass. Then, we use stoichiometry from the balanced equation to find the moles of O2 gas required. Finally, we apply the ideal gas law (PV=nRT) to find the volume of oxygen gas needed for the reaction.
Assuming the molar mass of KH is 39.1 g/mol (potassium's molar mass plus hydrogen's), we have:
Moles of KH = mass / molar mass = 70.25 g / 39.1 g/mol = 1.796 moles KH
The stoichiometry from the balanced equation shows that 2 moles of KH react with 1 mole of O2. Thus:
Moles of O2 = 1.796 moles KH / 2
Next, using the ideal gas law where R = 0.0821 L·atm/K·mol, T=298K (25°C+273), and P=1 atm:
Volume of O2 = (nRT) / P
By plugging the values into the ideal gas law equation, we can find the volume of O2 needed.