Question

A student performed an experiment in which hydrogen gas was collected over water from the reaction of magnesium and hydrochloric acid using the apparatus shown and accoridng to the reaction equation and the data collected shown below:

Mass of Mg	0.100 g
Volume of HCl	10 mL
Volume of H2 collected	57.5 mL
Temperature of H2 collected	22.0 oC
Barometric pressure	29.94 inHg

Mg(s) + 2 HCl(aq) → H₂(g) + MgCl₂(aq)                

a. Barometric pressure in atmospheres?

b. Water vapor pressure in atmospheres?

c. Pressure of the hydrogen in atmospheres?

d. Volume of hydrogen in liters?

e. Temperature in Kelvin?

f. Moles of hydrogen collected?

g. Moles of hydrochloric acid reacted?

h. Molarity of hydrochloric acid?

i. If the bottle claimed that the molarity was 0.510 M, what is the percent error?

Answer 1

The given chemical reaction is

Mg + 2 HCl -------> MgCl₂ + H₂

As per the balanced stoichiometric equation,

1 mole Mg = 2 moles HCl = 1 mole H₂

a) The barometric pressure at 22⁰C is given as 29.94 in Hg

The barometric pressure in atmospheres is P_bar = (29.94 in Hg)(25.4 mm/1 in)*(1 atm/760 mm Hg) = 1.0006 atm (since 1 in = 25.4 mm and 1 atm = 760 mm Hg) (ans).

b) The vapor pressure of water at 22⁰C can be easily obtained from tabulated values. Such tables are easily available in chemical databases. The value of the vapor pressure of water is P_H2O = 19.8 torr = 19.8 mm Hg (since 1 torr = 1 mm Hg) = (19.8 mm Hg)*(1 atm/760 mm Hg) = 0.0260 atm (ans).

c) The pressure due to dry hydrogen gas is given by P_H2 = P_bar – P_H2O = (1.0006 – 0.0260) atm = 0.9746 atm (ans).

d) The volume of hydrogen gas collected is 57.5 mL; the volume of hydrogen gas collected in liters is V = (57.5 mL)*(1 L/1000 mL) = 0.0575 L (ans).

e) The temperature of hydrogen gas collected is 22⁰C; the temperature in the Kelvin scale is T = (273 + 22) K = 295 K (ans).

f) We shall use the ideal gas law to compute the moles of hydrogen collected. The ideal gas law is

P*V = n*R*T where n = number of moles and R = 0.082 L-atm/mol.K

(0.9746 atm)*(0.0575 L) = n*(0.082 L-atm/mol.K)*(295 K)

===> n = 2.3167*10^-3 mole ≈ 2.317*10^-3 mole

The moles of hydrogen gas evolved = 2.317*10^-3 mole (ans).

g) Moles of hydrochloric acid reacted = (moles of H₂)*(2 moles HCl/1 mole H₂) = (2.317*10^-3 mole H₂)*(2 moles HCl/1 mole H₂) = 4.634*10^-3 mole HCl (ans).

h) Molarity of HCl = (moles of HCl)/(volume of HCl in L) = (4.634*10^-3 mole)/[(10 mL)*(1 L/1000 mL)] = 0.4634 mol/L = 0.4634 M (ans).

i) The claimed molarity is 0.510 M whereas the actual molarity obtained is 0.4634 M. The percent error in molarity is

Percent error = [(0.510 – 0.4634)M/(0.510 M)]*100 = 2.3766% ≈ 2.377% (ans).