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determine the pressure in kpa of hydrogen gas produced when 29.51g of aluminum reacts with excess sodium hydroxide and water if the temperature is 25.67c and the volume is 14.75L?

Answer :

tanisharawat014

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 .

Learn more about hydrogen gas here :

brainly.com/question/19813237

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