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 (0.001277 moles), calculate the average percent yield of H2 in your reaction
To calculate the average percent yield of H2 in your reaction, we first need to determine the maximum amount of H2 that could have been produced, which is based on the stoichiometry of the reaction.
From the balanced net ionic equation: Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g), we can see that 1 mole of Mg reacts with 2 moles of H+ to produce 1 mole of H2.
Based on the given starting amount of metal (0.052g), we need to convert this to moles of Mg. The molar mass of Mg is 24.31 g/mol.
Moles of Mg = 0.052g / 24.31 g/mol = 0.002138 moles of Mg
From the stoichiometry of the reaction, we can determine the maximum amount of H2 that could be produced, which is 0.002138 moles of Mg multiplied by the mole ratio of H2 to Mg, which is 1:2.
Maximum moles of H2 = 0.002138 moles of Mg * (1 mole of H2 / 1 mole of Mg) = 0.002138 moles of H2
Given that the amount of H2(g) produced in the reaction is 0.001277 moles, we can now calculate the percent yield.
Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (0.001277 moles of H2 / 0.002138 moles of H2) × 100 = 59.78%
Therefore, the average percent yield of H2 in your reaction is approximately 59.78%.