Question

What volume of o2 at 798 mmhg and 33 degree celsius is required to synthesize 22 mol of no?

Answer 1

To find the volume of O2 required to synthesize 22 mol of NO, we use the ideal gas law and the balanced reaction. After converting units and calculating, approximately 276 L of O2 at 798 mmHg and 33°C is needed.

Step-by-step explanation:

The question asks for the volume of O2 (oxygen gas) needed to produce 22 moles of NO (nitric oxide) gas. The given conditions are a pressure of 798 mmHg and a temperature of 33 degrees Celsius. To answer this, we can use the ideal gas law, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to balance the chemical equation for the reaction:

• N₂(g) + O₂(g) → 2 NO(g).

According to the stoichiometry of the reaction, 1 mole of N₂ reacts with 1 mole of O₂ to produce 2 moles of NO. Therefore, to produce 22 moles of NO, we need 11 moles of O₂.

Next, we convert the temperature to Kelvin:

• T = 33 + 273.15 = 306.15 K

Since pressure must be in atm for R = 0.0821 L·atm/mol·K, we convert the given pressure in mmHg to atm:

• P = 798 mmHg × (1 atm / 760 mmHg) = 1.05 atm

We can now rearrange the ideal gas law to solve for the volume (V) of O₂:

• V = (nRT) / P
• V = (11 mol × 0.0821 L·atm/mol·K × 306.15 K) / 1.05 atm
• V ≅ 276 L (rounded to three significant figures)

Therefore, approximately 276 L of O₂ are required to synthesize 22 mol of NO under the given conditions.