Question

Pre-lab for Molar Volume of a Gas How much volume does a gas occupy under standard conditions of temperature (O°C or 273 K) and pressure (1 atm or 760 mmHg)? As you knew from high school, after reading the lab manual, or from lecture, you already know the answer is 22.4 L per mole (L/mol). Thus, this is more of a verification than a true experiment. In a nutshell, you will generate hydrogen gas via a reaction, collect that gas over water (thus you will actually collect hydrogen and water vapor), do a calculation to subtract out the water, and then do a few additional calculations to get (a) the volume of pure hydrogen at standard T and P (a.k.a. STP) and (b) the moles of hydrogen gas collected. Assume you react 0.0279 g of Mg with HCle) until all of the Mg has reacted. Using the gas-collection tube (a.k.a eudiometer), you obtain 29.60 mL of gas in lab where the temperature is 20.1°C and the barometric pressure reads 744.8 mmHg. Use these data to calculate the molar volume of the gas according to steps 2-5 on page 42 of the lab manual. Please ignore step 1 and the part of equation (2) on page 41 for the time being (i.e. assume PH200- 0). We will address these in lab. Note that you willl need to look up the vapor pressure of water on in your textbook (look for a table of vapor pressures) for one of the calculations.

Answer 1

Mg + 2 HCl $\rightarrow$ MgCl₂ + H₂.

Number of moles of H₂ = number of moles of Mg reacted = 0.0279g/24.305g/mol = 0.0011479mol

Assuming P_H2O = 0

V = nRT/P = 0.0011479mol×0.0821atm-L/K.mol × 293.25K ÷(744.8/760)atm = 0.0282 L.  

Molar volume of gas = 0.0282L/0.0011479mol = 24.57L/mol

----------

At 20.1℃ , P_H2O = 17.582 mmHg

Pressure of H₂ = 744.8 - 17.582 = 727.22mmHg = 727.22/760 atm = 0.957atm

V = 0.0011479×0.0821×293.25/0.957 = 0.0289L

Molar volume of gas = 0.0289L/0.0011479mol = 25.16 L/mol