1 Aluminum reacts with aqueous sodium hydroxide to produce h

1) Aluminum reacts with aqueous sodium hydroxide to produce hydrogen gas according to the following equation:

2Al(s) + 2NaOH(aq) + 6H₂O(l) =====> 2NaAl(OH)₄(aq) + 3H₂(g)

The product gas, H₂, is collected over water at a temperature of 25 °C and a pressure of 758 mm Hg. If the wet H₂ gas formed occupies a volume of 6.59L, the number of moles of Al reacted was _____ mol. The vapor pressure of water is 23.8 mm Hg at 25 °C.

2) Oxygen gas can be prepared by heating potassium chlorate according to the following equation:

2KClO₃(s) ======> 2KCl(s) + 3O₂(g)

The product gas, O₂, is collected over water at a temperature of 25 °C and a pressure of 743 mm Hg. If the wet O₂ gas formed occupies a volume of 6.65L, the number of grams of O₂ formed is _____ g. The vapor pressure of water is 23.8 mm Hg at 25 °C.

Solution

Balanced equation:
2 Al(s) + 2 NaOH(aq) + 6 H₂O(l) ====> 2 NaAl(OH)₄(aq) + 3 H₂(g)

PV= nRT

P = Pressure in atm                       V= Volume in Liter

n = no of moles               R = 0.0821 L atm K-1 Mol-1

T = Temperature in Kelvin

Using the above formula we can calculate the moles

V=  6.59L

P = 758 -23.8 / 760 = 0.966 atm

T = 298 K

n = 0.966 x 6.59 / 0.0821 x 298 = 0.26 Moles

Moles of H2 produced = 0.26 Moles

Moels of Al produced = 0.173 Moles

Mass of Al produced = 0.173 * 26.98 = 4.67 gm

Question 2

The same calculation and forumula can be used

Moles of O2 produced = 0.946 x 6.65 / 0.0821 x 298 = 0.2572 Moles

number of grams of O2 formed = 0.2572 x 32 = 8.23 gm of O2