Based on your starting amount of metal (0.052g) , the net ionic equation of the reaction (Mg(s) + 2H+ (aq) —> Mg2+(aq) + H2 (g)) , and the amount of H2 (g) produced (0.001179 moles), calculate the average percent yield of H2 in your reaction
To calculate the average percent yield of H2 in the reaction, we first need to calculate the theoretical yield of H2. The theoretical yield is the maximum amount of product that can be formed, assuming 100% conversion of reactants.
1 mole of Mg produces 1 mole of H2, according to the balanced equation.
The molar mass of Mg is 24.31 g/mol, so the number of moles of Mg in the reaction is:
0.052 g / 24.31 g/mol = 0.00214 moles
Since the stoichiometry of the reaction tells us that 1 mole of Mg produces 1 mole of H2, the theoretical yield of H2 is also 0.00214 moles.
Now we can calculate the percent yield.
Percent yield = (actual yield / theoretical yield) * 100
In this case, the actual yield of H2 is given as 0.001179 moles.
Percent yield = (0.001179 moles / 0.00214 moles) * 100 = 55.1%
Therefore, the average percent yield of H2 in your reaction is 55.1%.