Question

In the reaction between solid alumium and 6.0M sulfuric acid, what is the minimum amount of sulfuric acid necessary (limiting reactant) to produce 25.0g of gaseous hydrogen?

In the above reaction what is the percent tield if the lab technician got 20g of Hydrogen gas in the laboratory?

Thank you :)

Answer 1

2Al + 3H2SO4 ---------> Al2(SO4)3 + 3H2

Mass of H2 = 25.0 g

Moles of H2 = Mass / Molar mass = 25.0 g / 2 g/mol = 12.5 mol

From the chemical reaction, it is clear that 3 mol of H2 is produced from 3 mol H2SO4.

So, 12.5 mol H2 is produced from (12.5 mol H2) x (3 mol H2SO4 / 3 mol H2) = 12.5 mol H2

Given, 6.0 M H2SO4.

Molarity = Moles / Volume

Volume (in L) = 12.5 mol / 6.0 M = 2.1 L

2.1 L of 6.0 M H2SO4 is necessary to produce 25.0 g of gaseous hydrogen.

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PErcent yield = (Actual yield / Theoretical yield) x 100

= (20.0g / 25.0 g) x 100 = 80%