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

Answer:

a)0.981 atm

b)0.000787 moles of hydrogen gas.

c)25.4 g

d)4.5%

Explanation:

a) Partial pressure of hydrogen gas = total pressure of gas collected - pressure of water vapour

Partial pressure of hydrogen gas= 1.01 atm - 0.029 atm = 0.981 atm

Number of moles of hydrogen gas can be obtained from the ideal gas equation;

PV= nRT

P= pressure of dry hydrogen gas= 0.981 atm

V= volume of hydrogen gas= 19.6 ml or 0.0196 L

T= 24° + 273 = 297 K

R= 0.082 L.atm.K-1.mol-1

n= PV/RT= 0.981 × 0.0196 / 0.082 × 297

n= 0.0192/24.4

n= 0.000787 moles of hydrogen gas.

From the reaction equation;

1 mole of the metal yielded 1 mole of hydrogen gas hence 0.000787 moles of the metal will yield 0.000787 moles of hydrogen gas.

Molar mass of the metal is obtained from;

Number of moles = reacting mass / molar mass

Molar mass = reacting mass/ number of moles

Molar mass= 0.0200 g /0.000787 moles

Molar mass of metal= 25.4 g

% error in the molar mass of the metal = 25.4-24.3/24.3 × 100/1

% error in the molar mass of the metal = 4.5%