What volume of 0.3682 M H2SO4 solution is required to react with 0.4198 grams of Al (CN) 3 according to the reaction, 2 Al (CN) 3 + 3 H2SO4  6 HCN + Al2 (SO4) 3?

Question

Miguel Daugherty

Chemistry

9 September, 06:26

What volume of 0.3682 M H2SO4 solution is required to react with 0.4198 grams of Al (CN) 3 according to the reaction, 2 Al (CN) 3 + 3 H2SO4  6 HCN + Al2 (SO4) 3?

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Answers (1)

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Kellen Franco · Answer 1

According to this reaction, the ratio between Al (CN) 3 and H2SO4 is 2:3. Knowing the molarity of H2SO4, in order to find out the volume we need to find out the moles of H2SO4 first.

Molar mass of Al (CN) 3 is 26.98 + (12.01+14.01) * 3 = 105.04 g/mol. So the moles of Al (CN) 3 reacted = mass/molar mass = 0.4198g/105.04g/mol = 0.003997mol. According to the above ratio, the moles of H2SO4 = 0.003997mol*3/2 = 0.005995mole.

The volume of H2SO4 needed for the reaction = moles of H2SO4/molarity of H2SO4 = 0.005995mole/0.3682mol/L = 0.01628L = 16.28 mL