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Sagot :
Explanation
Given
2H2(g) + O2(g) ⇨ 2H2O(g)
Volume = 6.00 L
At STP: Temperature = 273K
Pressure = 1 atm
We know that R constant = 0.0821 L.atm/mol.K
Solution
Step 1: Calculate the number of moles of oxygen at STP
Use the ideal gas law equation:
PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature.
n = PV/RT
n = (1 atm x 6.00 L)/(0.0821 L.atm/mol.K x 273 K)
n = 0.27 mol
Step 2: Use the stoichiometry to find the moles of H2
The molar ratio between oxygen and hydrogen is 1:2
Therefore the moles of H2 = 0.27 x (2/1) = 0.535 mol
Step 3: Calculate the volume of H2
V = nRT/P
V = (0.535 mol x 0.0821 L.atm/mol.K x 273 K)/1 atm
V = 12.0 L
Answer
Volume of hydrogen = 12.0 L
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