Question

What mass of Al2O3 is produced when 19.3 L of O2 gas react with

Al at 416 K and 6.88 atm. 4Al(s)+3O2(g)⟶2Al2O3(g) First, use the
ideal gas equation to calculate the moles of O2.

Answer 1

To find the mass of Al2O3 produced, we calculate the moles of O2 with the ideal gas law, then apply stoichiometry to determine the moles of Al2O3, which we then convert to mass using molar mass.

Step-by-step explanation:

The question revolves around a chemical reaction in which aluminum (Al) reacts with oxygen gas (O2) to produce aluminum oxide (Al2O3). To find out the mass of Al2O3 produced, we first need to calculate the moles of O2 using the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

We are given the pressure (P = 6.88 atm) and volume (V = 19.3 L) of the O2 gas, as well as the temperature (T = 416 K). The ideal gas constant (R) is generally given as 0.0821 L⋅atm/(K⋅mol).

Using the ideal gas law, we can calculate the moles (n) of O2:
n = PV / RT = (6.88 atm) × (19.3 L) / (0.0821 L·atm/(K·mol)) × (416 K)

Once we have the number of moles of O2, we can use the stoichiometry of the balanced chemical equation (4Al(s) + 3O2(g) → 2Al2O3(s)) to find out the number of moles of Al2O3 produced. Lastly, we can convert moles of Al2O3 to mass using its molar mass.

Answer 2

To calculate the moles of O2 gas, we can use the ideal gas equation PV = nRT. Given the pressure, volume, and temperature, we can plug in these values to calculate the moles of O2 gas.

Step-by-step explanation:

To calculate the moles of O2 gas, we can use the ideal gas equation: PV = nRT.

Given:

• Pressure (P): 6.88 atm
• Volume (V): 19.3 L
• Temperature (T): 416 K

Rearranging the equation, we have n = PV / RT.

Plugging in the values, we get n = (6.88 atm) * (19.3 L) / (0.0821 L·atm/mol·K * 416 K).

Calculating this gives us the number of moles of O2 gas.