Question

As the magnesium reacts, the hydrogen gas produced is collected by water displacement at 23.0oC. The pressure of the gas in the collection tube is measured to be 749 torr. Given that the equilibrium vapor pressure of water is 21 torr at 23.0oC, calculate the pressure that the H2(g) produced in the reaction would have if it were dry.

Answer 1

Answer: The pressure that the $H_2(g)$ produced in the reaction would have if it were dry will be 728 torr

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_{H_2}+p_{H_2O}$

Given : $p_{total}$  =total pressure of gas = 749 torr

$p_{H_2}$ = partial pressure of hydrogen = ?

$p_{H_2O}$ = partial pressure of water = 21 torr

putting in the values we get:

$749=p_{H_2}+21$

$p_{H_2}=728torr$

Thus the pressure that the $H_2(g)$ produced in the reaction would have if it were dry will be 728 torr