Answer:
Explanation:
Hello there!
In this case, according to the given combustion reaction of octane, it is possible for us to perform the stoichiometric method in order to calculate the mass of octane that is required to consume 300.0 g of oxygen by considering the 2:25 mole ratio, and the molar masses of 114.22 g/mol and 32.00 g/mol respectively:
Regards!
Answer: There are now 2.07 moles of gas in the flask.
Explanation:
P= Pressure of the gas = 697 mmHg = 0.92 atm (760 mmHg= 1 atm)
V= Volume of gas = volume of container = ?
n = number of moles = 1.9
T = Temperature of the gas = 21°C=(21+273)K= 294 K (0°C = 273 K)
R= Value of gas constant = 0.0821 Latm\K mol
When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.
P= Pressure of the gas = 775 mmHg = 1.02 atm (760 mmHg= 1 atm)
V= Volume of gas = volume of container = 49.8 L
n = number of moles = ?
T = Temperature of the gas = 26°C=(26+273)K= 299 K (0°C = 273 K)
R= Value of gas constant = 0.0821 Latm\K mol
Thus the now the container contains 2.07 moles.