[rudiment274]

The alkali metals are so reactive that they react directly with water in the absence of acid. For example, potassium reacts with water as follows:

2 K(s) + 2 H₂O(l)  2 K⁺(aq) + 2 OH^-(aq) + H₂(g)

How many milliliters of hydrogen will be evolved over water when 4.52 g K reacts with an excess of H₂O(l)?

T= 23°C, P= 758 mm Hg, P H₂O= 21 mm Hg
______________________________________________

First, I found moles of H₂

4.52 g K x (1 mol K/ 39.0983 g K) x (1 mol H₂/2 mol K)

=0.0578 mol H₂

Next, I converted the pressure to atm

758 mm Hg x (1 atm/760 mm Hg)= 0.997 atm

Then, I converted Celsius to Kelvin

23°C + 273= 296 K

Last, to find the volume of H₂, I used the ideal gas law

V= n_H2 x R x T/ P_H2

(0.0578 mol H₂ x 0.082 Lxatm/molxK) x 296 K/ 0.997 atm

Here's where I get stuck: They provided that the pressure for H₂O is 21 mm Hg, but I don't know where that fits in. What did I do wrong? Thank you for your *delete me*

[Borek]

Hint: collected gas is saturated with the water vapor. Its partial pressure will be 21 mmHg. You need to account for that.

[rudiment274]

All right. By Dalton's Law,

P H₂= P_atm - P H₂O

= 758 mm Hg - 21 mm Hg = 737 mm Hg

737 mm Hg x (1 atm/760 mm Hg)= 0.9697 atm

Then I find L of H₂ from the Ideal Gas Law and convert to milliliters.

[rudiment274]

A student reacts 0.1212 g of an Mg/NaCl mixture with hydrochloric acid. The volume of hydrogen saturated with water vapor collected at 20.0°C was measured to be 42.7 mL. The column of water that remained in the buret was 10.3 cm high. If vapor pressure of water = 17.5 mm Hg at this temperature and the barometric pressure was 763 mm Hg, what is the % Mg in the sample?
_______________________________

Since Mg reacts with HCl (aq) to produce magnesium chloride and hydrogen gas, the balanced chemical equation is:

Mg (aq) + 2 HCl (aq)  MgCl₂ (aq) + H₂(g)

By Dalton's Law, the partial pressure of the hydrogen is:

P H₂= P_atm - P_H2O - P_col

To find the pressure due to a column of water, we take the height of the water times the density of water all over the density of mercury.

P_col(mm Hg) = (10.3 cm x 1.00 g/cm³)/ 13.53 g/cm³

= 0.761 mm Hg

Now, we can use Dalton's Law:

P H₂= 758 mm Hg - 17.5 mm Hg - 0.761 mm Hg

= 739.7 mm Hg x (1 atm/760 mm Hg)= 0.9724 atm

The volume of the gas is 42.7 mL
The temp. of the gas is 20°C + 273= 293 K

Finally, the moles of hydrogen gas from the Ideal Gas Law is:

n_H2= P_H2 x V/ RT

= 0.9724 atm x 42.7 mL / (0.082 Lxatm/molxK) x 293 K

= 1.73 mol H₂(g)

Then we use molar mass and mass percent to obtain the percent Mg in the sample.

Is my process correct? Is there anything I did wrong? Your feedback is very helpful. Thank you!

[Borek]

All right. By Dalton's Law,

P H₂= P_atm - P H₂O

= 758 mm Hg - 21 mm Hg = 737 mm Hg

737 mm Hg x (1 atm/760 mm Hg)= 0.9697 atm

Then I find L of H₂ from the Ideal Gas Law and convert to milliliters.

Logic looks OK to me.

[Arkcon]

rudiment:, I hope you don't mind my merging your two threads.  You're asking similar questions, and help for one may help you for another.

[rudiment274]

Oh, I need to convert the 10.3 cm to mm!

10.3 cm x (1 mm/0.1 cm)= 103 mm

[Borek]

It wouldn't hurt to also convert mL to L.