Aluminum reacts with aqueous sodium hydroxide to produce hydrogen gas according

Question

Aluminum reacts with aqueous sodium hydroxide to produce hydrogen gas according to the following equation: 2Al(s) + 2NaOH(aq) + 6H2O() 2NaAl(OH)4(aq) + 3H2(g) In an experiment, the H2 gas is collected over water in a vessel where the total pressure is 758 torr and the temperature is 20°C, at which temperature the vapor pressure of water is 17.5 torr. Under these conditions, the partial pressure of H2 is __torr. If the wet H2 gas formed occupies a volume of 9.60 L, the number of moles of H2 formed is mol.

Explanation / Answer

According to dalton's law total pressure of gas is equal to sum of partial pressure exerted by each individual gas

Ptotal = P1 + P2 + P3

Phyrogen = Ptotal  - PH2O

substitute value

PHydrogen = 758 - 17.5 = 740.5 torr

Pressure of H2 = 740.5 torr

Use ideal gas equation for calculation of mole of gas

Ideal gas equation

PV = nRT             where, P = atm pressure= 740.5 torr = 0.9743 atm,

V = volume in Liter = 9.60 L

n = number of mole = ?

R = 0.08205L atm mol-1 K-1 =Proportionality constant = gas constant,

T = Temperature in K = 200C = 273.15+ 20 = 293.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.9743 X9.60) / (0.08205X 293.15) = 0.38886 mole

0.38886 mole of H2 produced.

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