Question

What volume of O2 at 722 mmHg and 41 °C is required to synthesize 20.0 mol of NO?Express your answer to three significant figures and include the appropriate units.

Answer 1

1) Write the chemical equation.

$N_2+O_2\rightarrow2NO$

2) List the known and unknown quantities.

Sample: Oxygen (O2).

Pressure: 722 mmHg.

Temperature: 41 °C.

Sample: Nitric oxide (NO).

Amount of substance: 20.0 mol NO

3) Moles of oxygen to produce 20.0 mol NO.

The molar ratio between O2 and NO is 1 mol O2: 2 mol NO.

$mol\text{ }O_2=20.0\text{ }mol\text{ }NO\ast\frac{1\text{ }mol\text{ }O_2}{2\text{ }mol\text{ }NO}=10.0\text{ }mol\text{ }O_2$

4) Volume of oxygen required.

4.1- Convert mmHg to atm.

1 atm = 760 mmHg

$atm=722\text{ }mmHg\ast\frac{1\text{ }atm}{760\text{ }mmHg}=0.95\text{ }atm$

4.2- Convert °C to K.

$K=\degree C+273.15$$K=41\degree C+273.15=314.15K$

4.3- Set the equation (ideal gas equation)

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

4.4- Plug in the know quantities and solve for V (liters).

$(0.95\text{ }atm)(V)=(10.0\text{ }mol\text{ }O_2)(0.082057\text{ }L\ast atm\ast K^{-1}\ast mol^{-1})(314.15\text{ }K)$$V=\frac{(10.0\text{ }mol\text{ }O_2)(0.082057\text{ }L\ast atm\ast K^{-1}\ast mol^{-1})(314.15\text{ }K)}{0.95\text{ }atm}$$V=271.3495426\text{ }L$

The volume of oxygen (O2) required is 271 L.