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When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. 2Al(s)+6HCL(aq)->2AlCl3(aq) +3H2(g) What mass of Al(s) is required to produce 513.0 mL of H2(g) at STP?

Donald Ward

in Chemistry

1 answer

1 answer

James Washington on September 8, 2018

Answer : The mass of aluminum (Al) is 0.413 gSolution : Given the Volume of STP = 513 ml = 0.513 L ( 1 L = 1000 ml )Molar mass of aluminium = 26.98 g/moleFirst, we need to calculate the moles of .At STP, 1 mol occupies 22.4 L volumenow, 0.513 L gives moles of moles of = 0.0229 molesThe Net balance of the chemical reaction is the balanced chemical reaction, we come to the conclusion that2 moles of Aluminum (Al) produces 3 moles of hydrogen gasNow the number of moles of aluminum required in 0.0229 moles of hydrogen gas = = 0.0153 molesNow, we need to calculate the mass of aluminum.The mass of aluminum = number of moles × Molar mass = 0.0153 mol x 26.98 g/mol = 0.413 edel mass of aluminum required is to 0.413 g.

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