Question

A mixture of 0.200 g of H2, 1.00 g of N2, and 0.820 g of Ar is stored in a closed container at standard temperature and pressure (STP). What is the volume of the container, assuming that the gases exhibit ideal behavior?

A.45.3 L B.4.75 L
C.1.00 L
D.3.48 L
E.6.51 L

Answer 1

To find the container volume at STP, we first calculate the number of moles of each gas present and then multiply the total moles by the molar volume (22.4 L/mol). The calculations yield a total volume of 3.49 L, which rounds to 3.48 L.

Step-by-step explanation:

To find the volume of the container storing the mixture of H2, N2, and Ar at STP (standard temperature and pressure), we'll use the concept that at STP, one mole of any ideal gas occupies a volume of 22.4 L. First, we need to calculate the number of moles of each gas using their respective molecular weights (H2 = 2 g/mol, N2 = 28 g/mol, Ar = 39.95 g/mol) and then use the molar volume to find the total volume.

• Moles of H2: 0.200 g / 2 g/mol = 0.100 mol
• Moles of N2: 1.00 g / 28 g/mol = 0.0357 mol
• Moles of Ar: 0.820 g / 39.95 g/mol = 0.0205 mol

Add up the moles of each gas to get the total number of moles present in the container, which gives us 0.1562 mol. Now, multiply this total by the molar volume (22.4 L/mol) to find the total volume:

Total Volume = 0.1562 mol × 22.4 L/mol = 3.49 L, which can be rounded to 3.48 L as per the options given.