10.0 grams of carbon dioxide and 20.0 grams of oxygen are mixed in a 50.0 L container at 30.0 degrees celcius. What is the pressure of this mixture in mm Hg? I defined my variables. so i had Mass=30.0g Volume=50.0L Temperature=303 Kelvin pressure=x.
then i used PV*GFM=mRT.
(x)(50.0L)(60)=30.0G(62.4)(303). The correct answer is 322 mmHG. But i'm not getting that. can anyone help? thanks:)
To solve this problem, you are on the right track by using 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 ideal gas constant, and T is the temperature.
However, there are a few corrections and clarifications needed in your calculation. Here's a step-by-step explanation of how to solve the problem correctly:
1. Convert the given mass of carbon dioxide and oxygen into moles. To do this, divide the mass by the molar mass of each gas.
- For carbon dioxide (CO2):
Molar mass of CO2 = 12.01 g/mol (for carbon) + 2 * 16.00 g/mol (for oxygen)
moles of CO2 = 10.0 g / 44.01 g/mol
- For oxygen (O2):
Molar mass of O2 = 2 * 16.00 g/mol (for oxygen)
moles of O2 = 20.0 g / 32.00 g/mol
2. Calculate the total number of moles (n) by adding the moles of CO2 and O2.
3. Convert the given temperature from Celsius to Kelvin by adding 273.15.
T = 30.0°C + 273.15 = 303.15 K
4. Plug in the values of n, V, R, and T into the ideal gas law equation.
PV = nRT
P * 50.0 L = n * 0.0821 L·atm/mol·K * 303.15 K
P = (n * 0.0821 * 303.15) / 50.0
5. Substitute the value of n from step 2 and solve for P.
6. To convert the pressure from atm to mmHg, multiply the calculated pressure by 760 (since 1 atm = 760 mmHg).
By following these steps, you should find the correct pressure in mmHg (the given answer is 322 mmHg).