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put a y Aluminum in the excess H2SO4 solution, get 13.44 liters of H2 dktc and mg of Salt Calculate a and m?

Answer :

mickymike92

From the equation of the reaction and the volume of hydrogen gas produced, a = 10.8 g and m = 68.4 g

What mass of salt is produced when aluminium metal reacts with excess H2SO4 solution?

The equation of the reaction reaction of aluminium metal in H2SO4 solution is given as follows:

[tex]2 \: Al(s) + 3 \: H_{2}SO_{4} \rightarrow Al_{2}(SO_{4})_{3} + 3 \: H_{2}(g) \\ [/tex]

2 moles of aluminium produces 3 moles of hydrogen gas.

Moles of hydrogen gas in 13.44 liters =13.44/22.4 = 0.6 moles

Moles of aluminium required = 0.6 × 2/3 = 0.4 moles

Molar mass of aluminium = 27 g/mol

Mass of aluminium = 0.4 × 27 = 10.8 g

Moles of salt produced = 0.6 × 1/3 = 0.2 moles

Molar mass of salt = 342 g/mol

Mass of salt = 0.2 × 342 = 68.4 g

Therefore, a = 10.8 g and m = 68.4 g

Learn more about mass of salt at: https://brainly.com/question/8896558

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