Answer to Question #107095 in General Chemistry for Sam

Question

When 37.7 L of nitric oxide reacts with 31.5 L of oxygen at 363 K under a constant pressure of 1.252 atm, what is the theoretical yield (in g) of nitrogen dioxide?

Answer 1

When 37.7 L of nitric oxide reacts with 31.5 L of oxygen at 363 K under a constant pressure of 1.252 atm, what is the theoretical yield (in g) of nitrogen dioxide?

Solution:

The equation of the chemical reaction is:

2NO + O₂ = 2NO₂

First of all we need calculate the mass of nitric oxide using Mendeleev-Clapeyron's equation:

pV = m*R*T/M

where m(NO) = pV*M(NO)/(RT), where p is pressure, V - volume, M - molar mass, R - universal gas constant, T - temperature.

Thus:

m(NO) = 1.252*37.7*(14+16)/(0.082*363) = 47.57 (g)

Use the same calculations for O₂ molecule:

m(O₂) = pV*M(O₂)/(RT) = 1.252*31.5*32/(0.082*363) = 42.40 (g)

Using chemical reaction we can determine the mass of product NO₂:

_{47.57g Xg}

2NO + O₂ = 2NO₂

_{2*30 32}

where mass of oxygen molecule equals:

X = m(O₂) = 47.57*32/60 = 25.37 (g)

So mass of O₂ was given in excess.

That's why determine the mass of product NO₂ using the mass of first reactant:

_{47.57g Xg}

2NO + O₂ = 2NO₂

_{2*30 2*(14+16*2)}

X = m(NO₂) = 47.57*92/60 = 72.94 (g)

We can see that the theoretical yield of product equals 72.94 g.

Answer: m(NO₂) = 72.94 g.