Calcium metal and water produce hydrogen gas. If the volume of wet gas collected over water is 95.0 mL at 20 °C and 758 mm Hg, what is the volume of dry gas at STP conditions? (The vapor pressure of water at 20 °C is 17.5 mm Hg.)

Question

Calcium metal and water produce hydrogen gas. If the volume of wet gas collected over water is 95.0 mL at 20 °C and 758 mm Hg, what is the volume of dry gas at STP conditions? (The vapor pressure of water at 20 °C is 17.5 mm Hg.)

Answer 1

pressure dry gas=pressure wet gas-vaporpressurewater

So you know pressuredrygas

volumegas=95ml
pressuregas=758-17.5
temp: 20C

V2*Pstp/tempstp=volumegas*pressure/temp

where on the right side, it refers to the conditions immediately above.

Answer 2

To find the volume of dry gas at STP conditions, we need to consider the relationship between pressure, temperature, and volume using the ideal gas law. The ideal gas law equation is:

PV = nRT

Where:
P is the pressure
V is the volume
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature in Kelvin

First, let's convert the given conditions to the correct units.
Initial pressure (P): 758 mm Hg
Vapor pressure of water (Pvap): 17.5 mm Hg
Total pressure (Ptotal) = P - Pvap = 758 mm Hg - 17.5 mm Hg = 740.5 mm Hg
Temperature (T): 20 °C = 293 K

The next step is to determine the number of moles present in the wet gas. To do this, we use the ideal gas law equation and rearrange it to solve for the number of moles (n):

n = (PV) / (RT)

Now we can substitute the values into the equation:

n = (740.5 mm Hg * 95.0 mL) / (0.0821 L·atm/mol·K * 293 K)

Note that we convert the volume from milliliters to liters by dividing by 1000.

n = (740.5 mm Hg * 0.0950 L) / (0.0821 L·atm/mol·K * 293 K)

Simplifying the equation:

n = 0.07394 mol

Now we know the number of moles of wet gas. Since the balanced chemical equation states a 1:1 ratio between calcium and hydrogen gas, we can conclude that the number of moles of hydrogen gas is also 0.07394 mol.

To find the volume of dry gas at STP, we can use the ideal gas law equation again:

P1V1 / T1 = P2V2 / T2

We know the initial conditions (P1 = Ptotal, V1 = 95.0 mL, and T1 = 293 K), and we want to find the volume at STP conditions (P2 = 1 atm and T2 = 273 K).

Converting the initial volume to liters:

V1 = 95.0 mL / 1000 = 0.0950 L

Now we can solve for V2:

(P1V1) / T1 = (P2V2) / T2

(740.5 mm Hg * 0.0950 L) / 293 K = (1 atm * V2) / 273 K

V2 = (740.5 mm Hg * 0.0950 L * 273 K) / (1 atm * 293 K)

Simplifying the equation:

V2 = 0.0794 L

Therefore, the volume of dry gas at STP conditions is 0.0794 liters.