Determine how many moles of water vapor will be produced at 1.00 atm and 200.oC by the complete combustion
of 10.5 L of methane gas (CH4)
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To determine the number of moles of water vapor produced by the complete combustion of methane gas, we need to use the balanced chemical equation for the combustion reaction. The combustion of methane can be represented by the equation:
CH4 + 2O2 → CO2 + 2H2O
From the equation, we can see that for every molecule of methane (CH4) consumed, 2 molecules of water (H2O) are produced.
Given that we have 10.5 L of methane gas, we need to convert the volume of methane gas into moles. To do this, we will use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, let's convert the temperature from Celsius to Kelvin.
T(K) = T(C) + 273.15
T(K) = 200 + 273.15
T(K) = 473.15 K
Now, we can calculate the number of moles of methane gas using the ideal gas law.
n = PV / RT
Assuming the pressure remains constant at 1.00 atm, we'll substitute the given values into the equation:
n = (1.00 atm * 10.5 L) / (0.0821 L·atm/(mol·K) * 473.15 K)
By calculating this expression, we find the number of moles of methane gas.
Next, since we have a balanced equation and know the stoichiometry, we can determine the number of moles of water vapor produced. From the balanced equation, we know that 1 mole of methane produces 2 moles of water vapor.
Finally, we can multiply the number of moles of methane gas by the stoichiometric ratio to find the number of moles of water vapor produced.
I'll leave the calculations to you.