10g of hydrogen and 64g of oxygen were filled in a steel vessel and exploded.vol. of gaseous product obtained
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2H2 + O2 ———–> 2H2O
10g of H2 = 10/2 = 5 mol
64g of O2 = 64/32 = 2 mol
O2 is limiting reagent.
At STP
22.4L of O2 gives 44.8L of H2O
44.8L of O2 will give 89.6L of H2O
Volume of gaseous product obtained is 89.6L
10g of H2 = 10/2 = 5 mol
64g of O2 = 64/32 = 2 mol
O2 is limiting reagent.
At STP
22.4L of O2 gives 44.8L of H2O
44.8L of O2 will give 89.6L of H2O
Volume of gaseous product obtained is 89.6L
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