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When hydrogen burns, water vapor is produced. the equation is 2h2(g) + o2(g) → 2h2o(g). if 12 l of oxygen are consumed at stp, what volume of water vapor is produced?

When hydrogen burns, water vapor is produced. the equation is 2h2(g) + o2(g) → 2h2o(g). if 12 l of oxygen are consumed at stp, what volume of water vapor is produced?
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When hydrogen burns, water vapor is produced. the equation is 2h2(g) + o2(g) → 2h2o(g). if 12 l of oxygen are consumed at stp, what volume of water vapor is produced?

User ShP
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1 Answer

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The volume of water vapor produced if 12 l of oxygen are consumed at STP is calculated as follows

2H2(g) +O2 (g) = 2H2O

by use of mole ratio between O2 to H2O which is 1:2 the number of liters of H2O produced is therefore

= 2/1 x12 L = 24 L
User REMESQ
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