A mixture of 0.224 g of H2, 1.06 g of N2, and 0.834 g of Ar is stored in a closed container at STP. Find the volume (in L) of the container,

Question

A mixture of 0.224 g of H2, 1.06 g of N2, and 0.834 g of Ar is stored in a closed container at STP. Find the volume (in L) of the container, assuming that the gases exhibit ideal behavior.

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Acacia 6 months 2021-07-28T03:03:18+00:00 1 Answers 10 views 0

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    2021-07-28T03:05:03+00:00

    Answer: The volume of given container is 3.83 L.

    Explanation:

    Given: Mass of H_{2} = 0.224 g

    Mass of N_{2} = 1.06 g

    Mass of Ar = 0.834 g

    Since, moles is the mass of a substance divided by its molar mass. Therefore, moles of given substances present in the mixture are as follows.

    Moles of  H_{2} are:

    Moles = \frac{mass}{molar mass}\\= \frac{0.224 g}{2 g/mol}\\= 0.112 mol

    Moles of N_{2} are:

    Moles = \frac{mass}{molar mass}\\= \frac{1.06 g}{28 g/mol}\\= 0.038 mol

    Moles of Ar are:

    Moles = \frac{mass}{molar mass}\\= \frac{0.834 g}{40 g/mol}\\= 0.021 mol

    Total moles = (0.112 + 0.038 + 0.021) mol = 0.171 mol

    Now, using ideal gas equation the volume is calculated as follows.

    PV = nRT

    where,

    P = pressure

    V = volume

    n = no. of moles

    R = gas constant = 0.0821 L atm/mol K

    T = temperature

    Substitute the values into above formula as follows.

    V = \frac{nRT}{P}\\= \frac{0.171 mol \times 0.0821 L atm/mol K \times 273 K}{1 atm}\\= 3.83 L

    Thus, we can conclude that the volume of given container is 3.83 L.

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