Hydrogen gas can produced by the reaction between metallic aluminum and aqueous hydrochloric acid

2Al+6HCl----2AlCl3+3H2

Hydrogen gas produced by this reaction is typically collected via water displacement, during which time the hydrogen gas becomes saturated with water vapor. If 271.6 mL of gas with a total pressure 1.03 atm was collected via water displacement at 29.4 °C, what is the partial pressure of hydrogen gas in the sample? How many grams of aluminum must have reacted to produce this quantity of hydrogen gas? The vapor pressure of water at 29.4 °C is 30.75 torr.

sorry but i just got soo confused

Ptotal = pH2(gas) + pH2O(gas)

You know Ptotal and pH2O, what is pH2.
Then PV = nRT tells you n for mols H2. From there you have a stoichiometry problem to go from mols H2 to mols Al to g Al.

Well, well, well! Let's break this down, shall we?

First, we need to find the partial pressure of hydrogen gas in the sample. Since the total pressure is 1.03 atm and the vapor pressure of water at 29.4°C is 30.75 torr, we need to subtract the vapor pressure of water from the total pressure to determine the partial pressure of hydrogen gas.

So, let's convert the vapor pressure of water from torr to atm. We have 30.75 torr, which is approximately 0.0406 atm.

Now, we can subtract the vapor pressure of water from the total pressure:

Partial pressure of hydrogen gas = Total pressure - Vapor pressure of water
= 1.03 atm - 0.0406 atm
= 0.9894 atm

So, the partial pressure of hydrogen gas in the sample is approximately 0.9894 atm.

Next, let's find the number of moles of hydrogen gas produced by using the Ideal Gas Law equation: PV = nRT. We have the volume (271.6 mL, convert to liters), the pressure (0.9894 atm), the temperature (29.4°C, convert to Kelvin), and the gas constant (0.0821 L·atm/mol·K).

Let's solve for moles of hydrogen gas:

n = PV / RT

Now, let's plug in the values:

n = (0.9894 atm) * (0.2716 L) / [(0.0821 L·atm/mol·K) * (29.4°C + 273.15)]

Put on your calculation hat and find the number of moles of hydrogen gas.

Finally, we need to determine the moles of aluminum (Al) that reacted to produce this quantity of hydrogen gas. According to the balanced chemical equation:

2Al + 6HCl → 2AlCl3 + 3H2

We can see that 2 moles of aluminum produce 3 moles of hydrogen gas. So, we set up a ratio:

2 moles Al / 3 moles H2 = x moles Al / moles of H2 calculated

Now, you need to solve for x, which represents the moles of aluminum. Once you have the moles of aluminum, you can use the molar mass of aluminum to calculate the mass in grams.

Voila! With this information, you can determine the partial pressure of hydrogen gas and the grams of aluminum that reacted. Don't forget to double-check your calculations, or you might end up with hilariously wrong results!

To find the partial pressure of hydrogen gas in the sample, we need to apply Dalton's Law of Partial Pressures. According to this law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases.

In the given reaction, the stoichiometry shows that 2 moles of aluminum react to produce 3 moles of hydrogen gas. We can use this information to calculate the number of moles of hydrogen gas produced.

First, we need to convert the given volume of gas collected to moles. To do this, we can use the ideal gas law equation:

PV = nRT

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

Given:
Total pressure (P) = 1.03 atm
Volume (V) = 271.6 mL (convert to liters by dividing by 1000) = 0.2716 L
Temperature (T) = 29.4 °C (convert to Kelvin by adding 273.15) = 302.55 K

Plugging these values into the equation, we can solve for the number of moles of the gas collected:

n = (P * V) / (R * T)
n = (1.03 atm * 0.2716 L) / (0.0821 L·atm/(K·mol) * 302.55 K)

The number of moles of gas collected is approximately 0.0111 mol.

Since there is a 2:3 mole ratio between aluminum and hydrogen gas, we can calculate the number of moles of aluminum that reacted:

0.0111 mol H2 * (2 mol Al / 3 mol H2) = 0.0074 mol Al

Next, we can use the molar mass of aluminum (26.98 g/mol) to calculate the mass of aluminum:

Mass = moles * molar mass
Mass = 0.0074 mol Al * 26.98 g/mol

The mass of aluminum that reacted to produce this quantity of hydrogen gas is approximately 0.199 g.

Finally, to find the partial pressure of hydrogen gas in the sample, we can rearrange the ideal gas law equation to solve for the partial pressure:

P = (n * R * T) / V

Using the same values as before but substituting the number of moles of hydrogen gas (0.0111 mol) and the volume (0.2716 L), we can calculate the partial pressure:

P = (0.0111 mol * 0.0821 L·atm/(K·mol) * 302.55 K) / 0.2716 L

The partial pressure of hydrogen gas in the sample is approximately 0.975 atm.

Therefore, the partial pressure of hydrogen gas in the sample is 0.975 atm and the mass of aluminum that reacted to produce this quantity of hydrogen gas is approximately 0.199 g.

Hydrogen gas produced by this reaction is typically collected via water displacement, during which time the hydrogen gas becomes saturated with water vapor. If 220.3 mL of gas with a total pressure 1.17 atm was collected via water displacement at 29.4 °C, what is the partial pressure of hydrogen gas in the sample?

How many grams of aluminum must have reacted to produce this quantity of hydrogen gas? The vapor pressure of water at 29.4 °C is 30.75 torr.

use the given info to find water vapor pressure of water at 29.4C, then subtract that from 1.03 atm.