2Al(s) + 6H2O(l) + 2OH-(aq) -> 2Al(OH)4- + 3H2(g)
If the temperature was 25oC and the vapor pressure of the hydrogen gas produced was 734.8 mm Hg, what was the volume of hydrogen gas generated?
First, we need to get the moles and molarity of each of the reactants:
`49.92 grams NaOH * (1 mol e NaOH)/(39.9971 grams NaOH) = 1.248 mol es NaOH = mol es OH^-`
Molarity of NaOH solution: (1.248 mol es)/(0.600L) = 2.08 M
`41.28 grams Al * (1 mol e Al)/(26.981 grams Al) = 1.530 mol es Al`
Next we need to determine the limiting reactant so we can be able to know how many moles of `H_2` gas are produced.
`2Al(s) + 6H_2O(l) + 2OH^(-)(aq) -> 2Al(OH)^(4-) + 3H_2(g)`
`1.248 mol es OH^(-) * (3 mol es H_2)/(2 mol esOH^(-))`
= 1.872 mol es `H_2`
`1.530 mol es Al * (3 mol es H_2)/(2 mol es Al)`
= 2.295 mol es `H_2`
From the results, we can see that the limiting reactant is the NaOH solution and therefore we will use the value derived from the NaOH solution. To determine the volume of the hydrogen gas that evolved from the reaction, we will use the ideal gas law, PV = nRT.
`PV = nRT`
- P = 734.8 mmHg/760 mmHg = 0.9668 atm
- n = 1.872 mol es H_2
- R = 0.08206 atm-L/mol-K
- T = 25 + 273.15 = 298.15K
`V = (nRT)/(P)`
`V = (1.872*0.08206*298.15)/(0.9668)`
V = 47.4 L -> answer