Answer :
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
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