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What volume of O2 at 836 mmHg and 21 ∘C is required to synthesize 14.5 mol of NO ?

User Robinkunde
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To find the volume of oxygen needed to synthesize 14.5 mol of nitrogen monoxide at 836 mmHg and 21 °C, we can use the ideal gas law equation.

Using the ideal gas law equation 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, we can calculate the volume of oxygen needed.

Given:

Pressure (P) = 836 mmHg (convert to atm: 1 atm = 760 mmHg)

Temperature (T) = 21 °C (convert to Kelvin: T(K) = T(°C) + 273.15)

Moles of NO (n) = 14.5 mol

First, convert the pressure to atm: 836 mmHg / 760 mmHg/atm = 1.099 atm

Next, convert the temperature to Kelvin: 21 °C + 273.15 = 294.15 K

Now, use the ideal gas law to find the volume (V) of oxygen: V = nRT / P

Substituting the given values:

V = (14.5 mol * 0.0821 L·atm/(mol·K) * 294.15 K) / 1.099 atm

Simplifying the calculation, the volume of oxygen required is approximately 344.77 L.

Therefore, to synthesize 14.5 mol of nitrogen monoxide at 836 mmHg and 21 °C, approximately 344.77 liters of oxygen are needed.

User Praneeth
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