Question

How many liters of O2 gas at 25°C and 1.00 atm pressure are needed to react with 60.25 grams of potassium hydride according to the chemical equation shown below?

2 KH(s) + O₂(g) → H₂O(l) + K₂O(s)

Answer 1

To find the volume of O2 needed, calculate the moles of KH from its mass, use the balanced equation to find the moles of O2 required, and then use the ideal gas law to determine the volume. 18.37 liters of O2 are needed to react with 60.25 grams of KH.

Step-by-step explanation:

Calculating the Volume of Oxygen Needed

To calculate the volume of O2 gas needed to react with 60.25 grams of potassium hydride (KH), we need to follow these steps:

1. Determine the molar mass of KH and convert the mass of KH to moles.
2. Use the balanced chemical equation to find the stoichiometric ratio between KH and O2.
3. Apply the ideal gas law to calculate the volume of O2 at the given conditions of temperature and pressure.

Firstly, 1 mole of KH weighs approximately 39.10 (K) + 1.01 (H) = 40.11 grams/mol. Thus, 60.25 grams of KH is equivalent to 60.25 g / 40.11 g/mol = 1.502 moles of KH.

According to the equation, 2 moles of KH react with 1 mole of O2. Therefore, 1.502 moles of KH will require 0.751 moles of O2.

To find the volume of O2, we use the ideal gas law: V = (nRT)/P. Here, R is the gas constant (0.0821 L atm/mol K), T is the temperature in Kelvin (25°C + 273 = 298 K), and P is the pressure in atm (1.00 atm).

The volume of O2 required is V = (0.751 mol * 0.0821 L atm/mol K * 298 K) / 1.00 atm = 18.37 liters.

Therefore, 18.37 liters of O2 at 25°C and 1.00 atm pressure are required to react with 60.25 grams of potassium hydride.