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17 October, 22:01

Oxygen gas reacts with powdered aluminum according to the following reaction:

4Al (s) + 3O2 (g) →2Al2O3 (s)

What volume of O2 gas, measured at 787 mmHg and 21 ∘C, is required to completely react with 55.0 g of Al?

Express the volume in liters to three significant figures.

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  1. 17 October, 22:08
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    35.5L of O2

    Explanation:

    4Al (s) + 3O2 (g) → 2Al2O3 (s)

    First, let us calculate the number of mole O2 that reacted with 55g of Al. This is illustrated below:

    Molar Mass of Al = 27g/mol

    Mass of Al = 55g

    Number of mole = Mass / Molar Mass

    Number of mole of Al = 55/27 = 2.04moles

    From the equation,

    4moles of Al reacted completely with 3 moles of O2.

    Therefore, 2.04moles of Al will react completely with = (2.04 x 3) / 4 = 1.53moles of O2

    Now let us obtain the volume of O2 occupied by 1.53moles of O2 measured at 787 mmHg and 21°C. This can be achieved by doing the following:

    n = 1.53moles

    P = 787mmHg

    Recall: 760mmHg = 1atm

    787mmHg = 787/760 = 1.04atm

    T = 21°C = 21 + 273 = 294K

    R = 0.082atm. L/Kmol

    V = ?

    Using the Ideal gas equation PV = nRT, we can obtain the volume as follows:

    PV = nRT

    V = nRT/P

    V = (1.53 x 0.082 x 294) / 1.04

    V = 35.5L

    Therefore, 35.5L of O2 required to react with 55.0g of Al
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