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?
Please explain how you got your answer
- Roger the MoleLv 71 month agoFavorite Answer
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.)