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11 August, 08:25

Li2O (s) + H2O (g) →2LiOH (s) How many grams of Li2O must be carried on board to remove 70.0 kg of water?

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Lithium oxide is used aboard the space shuttle to remove water from the atmosphere according to the equation

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  1. 11 August, 08:59
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    We need 116.1 kg of Li2O

    Explanation:

    Step 1: Data given

    MAss of water = 70 kg = 70000 grams

    Molar mass H2O = 18.02 g/mol

    Molar mass of Li2O = 29.88 g/mol

    Step 2: the balanced equation

    Li2O (s) + H2O (g) →2LiOH (s)

    Step 3: Calculate moles H2O

    Moles H2O = mass H2O / molar mass H2O

    Moles H2O = 70000 grams / 18.02 g/mol

    Moles H2O = 3884.6 moles

    Step 4: Calculate moles Li2O

    For 1 mol Li2O we need 1 mol H2O to produce 2 moles LiOH

    For 3884.6 moles H2O we need 3884.6 moles Li2O

    Step 5: Calculate mass Li2O

    MAss Li2O = moles * molar mass

    MAss Li2O = 3884.6 moles * 29.88g/mol

    Mass Li2O = 116071.8 grams = 116.1 kg

    We need 116.1 kg of Li2O
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