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18 October, 15:05

How many grams of N2 gas are present in 1.13 L of gas at 2.09 atm and 291 K?

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  1. 18 October, 18:33
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    Mass = 2.77g

    Explanation:

    Applying

    P=2.09atm, V = 1.13L, R = 0.082, T = 291K, Mm of N2 = 28

    PV=nRT

    NB

    Moles (n) = m/M

    PV=m/M*RT

    m = PVM/RT

    Substitute and Simplify

    m = (2.09*1.13*28) / (0.082*291)

    m = 2.77g
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