What volume of carbon dioxide, at 1 atm pressure and 112°C, will be produced when 80.0 grams of methane is burned?
1 answer:
Answer:
158 L.
Explanation:
What is given?
Pressure (P) = 1 atm.
Temperature (T) = 112 °C + 273 = 385 K.
Mass of methane CH4 (g) = 80.0 g.
Molar mass of methane CH4 = 16 g/mol.
R constant = 0.0821 L*atm/mol*K.
What do we need? Volume (V).
Step-by-step solution:
To solve this problem, we have to use ideal gas law: the ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. The formula is:
Where P is pressure, V is volume, n is the number of moles, R is the constant and T is temperature.
So, let's find the number of moles that are in 80.0 g of methane using its molar mass. This conversion is:
So, in this case, n=5.
Now, let's solve for 'V' and replace the given values in the ideal gas law equation:
The volume would be 158 L.
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