Question

A small piece of Cr metal reacts with dilute HNO3 to form H2 (g), which is collected over water at 18 C in a large flask. The total pressure in the flask is 753 mmHg.
Determine the partial pressure of the H2 present.

Answer 1

Answer:

The partial pressure of the H₂ is 737.5 mmHg

Explanation:

The partial pressure of the H₂ is equal to:

$P_{total} =P_{H_{2} } +P_{H_{2}0 }$

Where PH₂O is the partial pressure for the water. Clearing the partial pressure of H₂:

$P_{H_{2} } =P_{total} -P_{H_{2}0 }=753-15.5=737.5mmHg$

Answer 2

Complete Question

The complete question is shown on the first uploaded image

Answer:

The partial pressure is $P_p= 737.5 mm \ of \ Hg$

Explanation:

The Partial pressure of $H_2$ is mathematically represented as

           $P_p = P_T -P_w$

Where $P_T$ is the total pressure of water with a value of 15.5 mm of Hg

           $P_w$ is the partial pressure of water with a value 753 mm of Hg

Now substituting values

          $P_p = 753-15.5$

               $P_p= 737.5 mm \ of \ Hg$