if 0.0334 g Mg reacts with 15.00 mL of 6.0 M HCL what is the theoretical yield of the hydrogen gas?

Answer 1

We need a balanced equation for the reaction happening:

Mg + 2HCl --> MgCl₂ + H₂

Then figure out the number of moles of the two reactants:

.0334g Mg/2 x 1 mol Mg/24.305g = 1.34 x 10^-3mol Mg

Since M = n/L 6=n/.015 and n (# of moles) = .09 mol HCl

Determine the limiting reactant. This reaction requires 2x the number of mols of HCl compared to Mg. H₂ from HCl would be .045 while from Mg would be 1.34 x 10^-3. Since the latter is a smaller value, that identifies Mg as the limiting reactant and that amount as the number of mols of H₂ that will form.

The post is not specific about what units the yield should be in. If it is liters, and we assume STP (where a gas is assumed to have 22.4L/mol) then 1.34 x 10^-3 mol H₂ x 22.4L/mol gives .031 L H₂. If it is grams, replace the 22.4 L/mol with 2.016 g H₂/mol and the result is 2.701 g H_2.