A 0.250g smaple of magnesium-aluminum alloy dissolves completely in an excess of HCl(aq). When the liberated H2(g) is collected over water at 29C at 752torr, the volume is found to be 297mL. The vapor pressure of water at 29C is 30.0torr.

What is the mass percentage of aluminum in this alloy?

*already calculated the moles of H2 which was 1.14x10^-2

Thanks!

I'll assume your value for n is correct.

You need to solve for grams Al and grams Mg.
Let X = grams Al
then Y = grams Mg
==================
X + Y = 0.250
X(2*molar mass Al/3*molar mass H2) + Y(molar mass Mg/molar mass H2) = moles H2.
Solve the two equations simultaneously for X and Y.
Then %Al = (grams Al/grams sample)*100 = ?

To find the mass percentage of aluminum in the alloy, you need to determine the amount of aluminum in the alloy.

First, let's calculate the moles of hydrogen gas (H2) using the ideal gas law equation:

PV = nRT

Where:
P = pressure of the gas (752 torr - 30.0 torr)
V = volume of the gas (297 mL or 0.297 L)
n = moles of the gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (29°C + 273.15 = 302.15 K)

So, plugging in the values:

(752 torr - 30.0 torr) × (0.297 L) = n × (0.0821 L·atm/mol·K) × (302.15 K)

Let's solve for n:

722 torr × 0.297 L = n × 24.911 L·torr/mol·K

214.634 torr·L = n × 24.911 L·torr/mol·K

n = 214.634 torr·L / 24.911 L·torr/mol·K

n = 8.61 mol

Since 1 mole of magnesium-aluminum alloy produces 1 mole of hydrogen gas, the number of moles of the alloy is also 8.61 mol.

Now, let's use the molar mass of aluminum (Al) and the moles of the alloy to calculate the mass of aluminum in the sample.

The molar mass of Al is 26.98 g/mol.

Mass of Al = moles of Al × molar mass of Al

Since the moles of the alloy is equal to the moles of Al in the alloy:

Mass of Al = 8.61 mol × 26.98 g/mol

Mass of Al = 232.48 g

Next, let's calculate the mass percentage of aluminum in the alloy:

Mass percentage of Al = (Mass of Al / Mass of alloy) × 100%

To find the mass of the alloy, you'll need to convert the mass of hydrogen gas to the mass of the alloy.

The molar mass of H2 is 2.02 g/mol.

Mass of H2 = moles of H2 × molar mass of H2

Mass of H2 = 8.61 mol × 2.02 g/mol

Mass of H2 = 17.4022 g

Since the sample completely dissolved in HCl, the mass of the alloy is equal to the mass of H2:

Mass of alloy = 17.4022 g

Now, let's calculate the mass percentage of aluminum in the alloy:

Mass percentage of Al = (Mass of Al / Mass of alloy) × 100%

Mass percentage of Al = (232.48 g / 17.4022 g) × 100%

Mass percentage of Al ≈ 1336.7%

Therefore, the mass percentage of aluminum in the magnesium-aluminum alloy is approximately 1336.7%.

Note: The result seems higher than expected, which could indicate an error in calculation or experimental data. I recommend double-checking the calculations and ensuring the experimental measurements are accurate.

To find the mass percentage of aluminum in the alloy, we need to determine the moles of aluminum and magnesium in the sample.

First, let's find the moles of hydrogen gas (H2) using the ideal gas law:

PV = nRT

where P is the total pressure (752 torr - 30 torr = 722 torr in this case), V is the volume of H2 gas (297 mL), n is the number of moles of H2, R is the ideal gas constant, and T is the temperature in Kelvin (29°C = 29 + 273 = 302 K).

R = 0.0821 L·atm/(mol·K) (ideal gas constant)
P = 722 torr
V = 297 mL = 0.297 L
T = 302 K

Plugging in these values into the ideal gas law equation, we can find the number of moles of H2:

(722 torr) · (0.297 L) = n · (0.0821 L·atm/(mol·K)) · (302 K)

n = (722 torr) · (0.297 L) / (0.0821 L·atm/(mol·K)) · (302 K)
n ≈ 0.0348 moles

Now, since we know that 1 mole of magnesium reacts with 1 mole of HCl to produce 1 mole of H2, we can conclude that the moles of magnesium in the alloy are also 0.0348.

To find the moles of aluminum, we need to subtract the moles of magnesium from the total moles of the alloy:

moles of aluminum = total moles of the alloy - moles of magnesium
moles of aluminum ≈ 0.0348 moles - 0.0348 moles
moles of aluminum ≈ 0 moles

This calculation implies that there is no aluminum present in the alloy.

Finally, to calculate the mass percentage of aluminum in the alloy, we use the formula:

Mass percentage of aluminum = (mass of aluminum / mass of alloy) × 100%

Since there is no moles of aluminum, the mass of aluminum is 0 g. The mass of the alloy is given as 0.250 g.

Mass percentage of aluminum = (0 g / 0.250 g) × 100%
Mass percentage of aluminum = 0%

Therefore, the mass percentage of aluminum in this alloy is 0%.