If 500.0 mL of hydrogen are collected over water at 20.0 C and 99.3 kPa what will the volume of the dry gas be at STP?
To find the volume of the dry gas at STP, we can use the concept of the ideal gas law. The ideal gas law equation is:
PV = nRT
In this equation:
P is the pressure of the gas (in this case, 99.3 kPa)
V is the volume of the gas (in this case, initially 500.0 mL)
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/(K·mol))
T is the temperature in Kelvin (20.0 C = 293.15 K)
To convert the volume of the gas from mL to L, we divide it by 1000:
V = 500.0 mL / 1000 = 0.5000 L
Now we can rearrange the ideal gas law equation to solve for the volume at STP (standard temperature and pressure), where the pressure is 1 atm and the temperature is 0°C (273.15 K).
V1 / T1 = V2 / T2
V1 is the initial volume (0.5000 L)
T1 is the initial temperature (293.15 K)
V2 is the volume at STP (what we are trying to find)
T2 is the temperature at STP (273.15 K)
Simplifying the equation:
(0.5000 L) / (293.15 K) = V2 / (273.15 K)
Cross multiplying:
V2 = (0.5000 L) * (273.15 K) / (293.15 K)
Calculating the volume at STP:
V2 = 0.463 L
Therefore, the volume of the dry gas at STP is 0.463 L.