Question

Using the answer from problem$8.61$, how many grams of nitrogen are in Whitney's lungs at STP if the air contains $78\%$nitrogen?

Short answer

Whitney's lungs are in$2.75gof{N}_2$.

Step-by-step solution

Given Information

We need to find how many grams of nitrogen are in Whitney's lungs.

Simplify

Now,

$2.8L$is $100\%$of air and it is equal to the lung capacity.

so,

$\frac{2.8L}{100\%}$

calculate $78\%$of $2.8L$:

$78\%\times \frac{2.8L}{100\%}=2.2L$

The volume of the pure nitrogen in the air mixture of gases is $2.2L$.

Conversion factor:

$\frac{1molof{N}_2}{22.4Lof{N}_2}$

calculating mole:

$\frac{1molof{N}_2}{22.4Lof{N}_2}\times 2.2Lof{N}_2=0.098molof{N}_2$

We can present this with the conversion factor below:

$\frac{1molof{N}_2}{22.4Lof{N}_2}$

Calculating how many grams of nitrogen :

$\frac{1molof{N}_2}{22.4Lof{N}_2}\times 0.098molof{N}_2=2.75gof{N}_2$