Assume that you add 50.0 L of H2 at 395 mm Hg and 45°C to 85.0 L of O2 at 650 mm Hg and 10°C. How many mol of H2O can be produced?
We know that , 1mmHg = (1/760)atm
R = gas constant = 0.082L-atm/(mol.K)
Using the ideal gas equation for calculating the Number of Mole of hydrogen and oxygen.
PV = nRT
And
n = PV/RT ......(1)
Fir hydrogen gas
P = 395mmHg = (395/760) atm = 0.5197atm
L = 50.0L
T = 45°C = (45+273)K = 318K
Putting the all value in equation number (1)
n = (0.5197atm)(50.0L)/{(0.082L-atm/mol.K)(381K)}
n = 0.9965mol
Number of mole of Hydrogen = 0.9965mol
For Oxygen gas
P = 650mmHg = (650/760)atm = 0.8552atm
V = 85.0L
T = 10°C = (10+273)K = 283K
Putting the all value in equation number (1)
n = (0.8552atm)(85.0L)/{(0.082L-atm/mol.K)(283K)}
n = 3.1327mol
Number of moles of nitrogen = 3.1327mol
Balance Chemical reaction between hydrogen and oxygen will be given as follows.
2H2 + O2 -------> 2H2O
From the stoichiometry of chemical reaction.
2mol Hydrogen react with 1 mol Oxygen and produced 2 mol H2O,
so,
1mol hydrogen react with (1/2) mol Oxygen and produced (2/2) mol H2O.
so,
0.9965 mol of Hydrogen react with (1/2)×0.9965mol of Oxygen and produced (2/2)×0.9965 mol H2O.
Hence ,
Number of moles of H2O = (2/2)×0.9965mol
Number of mole of H2O = 0.9965mol
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