Question

calculate the volume of hydrogen produced when 6g of magnesium is reacted with 100cm of 1 mol l-1 hydrochloric acid
no of moles of Mg= mass/gfm = 6/24.3 = 0.247 moles

1/2 mole of Mg= 1/2 mole of H2

= 0.247 moles x 100 = 24.7cm3 ???

Answer 1

This is a limiting reagent (LR) problem and we know that because amounts are given for BOTH reactants. I do these the long way.
mols Mg = your number is right at 0.247 and that will produce 0.247 mols H2 if you had all of the HCl needed.

mols HCl = M x L = 1 x 0.1 = 0.1
0.1 mols HCl x (1 mol H2/2 mols HCl) = 0.1 x 1/2 = 0.0.05 mols H2.
These two values for mols H2 don't agree; the answer in LR problems is ALWAYS the smaller value; therefore, you will produce 0.05 mols H2.
1 mol occupies 22.4 L at STP.
0.05 mols x 22.4 L/mol = ?
You don't list any conditions; you must mean at STP.

Answer 2

Thank you that was really helpful Dr Bob