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no of moles of H2 = 10/2 = 5
no of moles of oxygen = 64/32 = 2
Total no of moles = 7
PV =nRT
Substituting the given values
P = 2752700 N/m2
On ignition
2H2+O2--> 2H20
We have 5 moles of H2 and 2 moles of O2
Based on stichiometry 2 moles of O2 will combine with 4 moles of H2 to form 4 moles of H20
Thus 1 mole of hydrogen will be left
Thus no of moles in the vessel = 1(H2)+4(H2O)
PV = nRT
Substituting the values and n= 5 we get
P = 1966261 atm
no of moles of oxygen = 64/32 = 2
Total no of moles = 7
PV =nRT
Substituting the given values
P = 2752700 N/m2
On ignition
2H2+O2--> 2H20
We have 5 moles of H2 and 2 moles of O2
Based on stichiometry 2 moles of O2 will combine with 4 moles of H2 to form 4 moles of H20
Thus 1 mole of hydrogen will be left
Thus no of moles in the vessel = 1(H2)+4(H2O)
PV = nRT
Substituting the values and n= 5 we get
P = 1966261 atm
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