Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 month ago

A mixture of 512.0 g of O2(g) and 15.12 g of H2(g) is placed in a balloon at standard temperature and pressure. ?

A mixture of 512.0 g of O2(g) and 15.12 g of H2(g) is placed in a balloon at standard temperature and pressure. After the oxygen gas and hydrogen gas react to form water, what is the volume of the balloon?

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  • 1 month ago
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    2 H2 + O2 → 2 H2O

    (512.0 g O2) / (31.99886 g O2/mol) = 16.00 mol O2

    (15.12 g H2) / (2.01588 g H2/mol) = 7.500 mol H2

    7.500 moles of H2 would react completely with 7.500 x (1/2) = 3.750 moles of O2, but there is more O2 present than that, so O is in excess and H2 is the limiting reactant.

    After the reaction there is only water and O2 left in the balloon.

    Supposing the water vapor has condensed, making its volume negligible:

    (16.00 mol O2) - (3.750 mol O2 reacted) x (22.414 L/mol) = 275 L

    (If the water was still present as vapor, the volume of the balloon would be significantly larger.)

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