Q. 8.62 - An Introduction to General, Organic, and Biological Chemistry

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?

Step-by-Step Solution

Answer

Whitney's lungs are in  $2.75 g of N_2$.

Step 1: Given Information

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

Step 2: Simplify

Now,

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

so,

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

calculate $78\%$ of $2.8 L$:

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

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

Conversion factor:

$\frac{1mol of N_2}{22.4 L of N_2}$

calculating mole:

$\frac{1mol of N_2}{22.4 L of N_2}\times 2.2L of N_2=0.098 mol of N_2$

We can present this with the conversion factor  below:

$\frac{1mol of N_2}{22.4 L of N_2}$

Calculating how many grams of nitrogen :

$\frac{1mol of N_2}{22.4 L of N_2}\times 0.098 mol of N_2=2.75 g of N_2$