The pressure of hydrogen gas is 276.25 Kpa .
Given,
Mass of aluminum = 29.51g
Temperature (T) = 25.67 degC =298.67 K
Volume (V) = 14.75 L
The required equation when aluminum reacts with excess sodium hydroxide and water is given by ,
2Al + 2NaOH + 2H2O ==>2NaAlO2 + 3H2
molecular mass of aluminum =26.98 g
1 mole of aluminum= 26.98g
2 moles of aluminum = 53.96g
2 mole of aluminum produces = 3 moles of hydrogen gas
53.96 g of aluminum produces = 6 g of hydrogen gas
29.51 g of aluminum produces = 6*(29.51) /53.96 =3.28 g of hydrogen gas
Thus ,
2 g of hydrogen gas = 1 mole of hydrogen
3.28 g of hydrogen gas = 3.28/2 mole =1.64 mol of hydrogen
Thus , n = 1.64 mol
According to ideal gas equation ,
PV=nRT
P=nRT/V
P = 1.64 * (0.0821L atm K^-1mol^-1 ) *(298.67 K)/14.75L
P=2.726 atm
P= 276.25 Kpa
Hence the pressure of hydrogen gas is 276.25 Kpa .
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