22.4dm3 of volume of each H2 and O2 are sparked to produce water vapours(consider as ideal gas) on completion of reaction what is decrease in volume of vessel.
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1dm³=1L
Since both H2 and O2 occupies 22.4L at STP, then consider the volume of the vessel be 22.4L+22.4L=44.8L.
2 H2(g)+ O2(g)------> 2 H2O(g)
Here H2 is the limiting reagent, thus
2moles of H2(44.8L)------>2 moles of H2O
1mole of H2(22.4L)------>1mole of H2O i.e 22.4L is produced.
Volume of O2 left= 22.4L-11.2L=11.2L (they react in 2:1 ratio)
Now volume present in vessel=22.4L of H2O + 11.2L of O2 = 33.6L
Decrease in volume of vessel = 44.8L - 33.6L = 11.2L
Since both H2 and O2 occupies 22.4L at STP, then consider the volume of the vessel be 22.4L+22.4L=44.8L.
2 H2(g)+ O2(g)------> 2 H2O(g)
Here H2 is the limiting reagent, thus
2moles of H2(44.8L)------>2 moles of H2O
1mole of H2(22.4L)------>1mole of H2O i.e 22.4L is produced.
Volume of O2 left= 22.4L-11.2L=11.2L (they react in 2:1 ratio)
Now volume present in vessel=22.4L of H2O + 11.2L of O2 = 33.6L
Decrease in volume of vessel = 44.8L - 33.6L = 11.2L
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