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What volume of CO2(g) is needed to react with 10.25g of LiOH at 21 C and 781mmHg?

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CO2(g) + 2LiOH(s) ======> H2O(l) + Li2CO3(s)

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  1. Moles LiOH = 10.25 g / 23.9484 g/mol = 0.428

    moles CO2 needed = 0.428/2 = 0.214

    T = 21 + 273 =294 K

    p = 781/760 = 1.03 atm

    V = nRT/p = 0.214 x 0.0821 x 294 / 1.03 =5.01 L

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