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15 February, 10:17

How many liters of oxygen gas, at 285 K and 1.2 atm, will react with 35.4 grams of calcium metal? Show all of the work used to solve this problem.

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  1. 15 February, 13:16
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    1) Chemical equation:

    2Ca + O2 - - - > 2CaO

    2) molar ratios

    2 mol Ca : 1 mol O2 : 2 mol CaO

    3) Convert 35.4 g of Ca to moles:

    number of moles = mass in grams / atomic mass = 35.4 g / 40.078 g/mol = 0.883 mol

    3) use proportion and solve for x:

    2 mol Ca / 1 mol O2 = 0.883 mol Ca / x

    => x = 0.883 mol Ca * 1 mol O2 / 2 mol Ca

    => x = 0.4416 mol O2

    3) use ideal gases equation to transform moles to liters

    pV = nRT = > V = nRT / p =

    V = [0.4416 mol * 0.0821 atm*liter / atm*K * 285K] / 1.2 atm = 8.6 liter

    Answer: 8.6 liters
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