Final answer:
To calculate the pressure of a mixture containing 3.90 g of H2 and 7.400 g of N2 at 273°C in a 10.0 L container, convert the masses of each gas to moles, add up the total moles, and apply the Ideal Gas Law. The resultant pressure for this mixture is calculated to be 11.74 atm.
Step-by-step explanation:
The pressure exerted by a mixture of gases can be calculated using the Ideal Gas Law, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature in Kelvin.
To find the pressure exerted by the mixture of H2 and N2, we first need to convert the given masses of each gas to moles using their respective molar masses. Hydrogen (H2) has a molar mass of 2 g/mol and nitrogen (N2) has a molar mass of 28 g/mol.
Number of moles of H2 = 3.90 g / 2 g/mol = 1.95 mol
Number of moles of N2 = 7.400 g / 28 g/mol = 0.264 mol
Total moles of gas = 1.95 mol + 0.264 mol = 2.214 mol
Next, we convert the temperature from degrees Celsius to Kelvin: T = 273 + 273 = 546 K
Using the Ideal Gas Law:
P = (nRT) / V
Where R = 0.0821 L·atm/(mol·K) (universal gas constant), we calculate the total pressure:
P = (2.214 mol * 0.0821 L·atm/(mol·K) * 546 K) / 10.0 L
P = 11.74 atm
The pressure exerted by the mixture of 3.90 g of H2 and 7.400 g of N2 at 273°C in a 10.0 L container is 11.74 atm.