Question

A 0.0200 gram piece of unknown alkaline earth metal, M, is reacted with excess 0.500 M H​₂S​O₄,​ and the hydrogen gas produced is collected over water. The total gas pressure inside the collecting tube is 1.01 atm, the air temperature 24.0​ °​C. The volume of gas collected is 19.6 mL. The gas in the tube contains water vapor, at a pressure of 0.029 atm.

Equation: M(s) + H2SO4(aq) ---> MSO4(aq) + H2(g)

A) What is the partial pressure of the dry hydrogen gas collected in the tube?

B) How many moles of hydrogen gas were collected?

C) Use the data provided to calculate the molar mass of the unknown metal.

D) Metal M has an actual molar mass of 24.3 g/mol. Calculate the % error in the experimental molar mass.

Answer 1

A) 0.981 atm

B) 7.89 × 10⁻⁴ moles

C) 25.348 grams ≈ 25.3 grams

D) 4.315% increase error in molar mass

Step-by-step explanation:

A) The partial pressure of the water vapor = 0.029 atm.

Total gas pressure in the tube = 1.01 atm

Therefore, by Dalton's law of partial pressure, the total pressure of a given mass of gas is equal to the sum of the partial pressures of the individual gases in the mixture

Hence, total pressure = partial pressure of H₂ + Partial pressure of water vapor

1.01 atm = partial pressure of H₂ + 0.029 atm.

∴ Partial pressure of H₂ = 1.01 atm - 0.029 atm = 0.981 atm

The partial pressure of the dry hydrogen gas collected in the tube = 0.981 atm

B) From the universal gas equation, we have;

PV = nRT

Therefore,

`n = (P * V)/(R * T)`

Where:

n = Number of moles

P = Pressure = 0.981 atm

V = Volume = 19.6 ml

T = Temperature = 24° = (273.15 + 24) K = 297.15 K

R = Universal Gas Constant = 0.08205 L·atm/(mol·K)

Plugging in the values, we have;

`n = (0.981 * 19.6 * 10^(-3) )/(0.08205 * 297.15 ) = 7.89* 10^(-4) \ moles`

The number of moles of hydrogen gas collected = 7.89 × 10⁻⁴ moles

C) Since 1 mole of M reacts with 1 mole of H₂SO₄ in the reaction, we have

0.500 M of H₂SO₄ will react with 0.500 M of M

From;

`Number \ of \ moles = (Mass)/(Molar \ mass)` we have;

`Molar \ mass = (Mass)/( Number \ of \ moles) = (0.0200 \, grams)/( 7.89 * 10^(-4) \, moles) = 25.3485 \ grams`

D) The
`Percentage \ error = (Approximate \ value - Exact \ value )/(Exact \ value ) * 100 = (25.3485 - 24.3)/(24.3) = 4.315\%` is

Given that the actual molar mass = 24.3 g/mol.

The percentage error in the experimental molar ≈ 4.315%.