Based on your starting amount of metal (0.052g), the net ionic equation for the reaction (Mg(s) + 2H+ (aq) —> Mg2+ (aq) + H2 (g)), and the amount of H2 produced (0.001179 mole), calculate the average percent yield of H2 in your reaction
To calculate the average percent yield of H2 in the reaction, we need to compare the amount of H2 produced (0.001179 mole) to the theoretical amount of H2 that should have been produced based on the starting amount of metal.
1. Calculate the theoretical amount of H2 produced:
- From the balanced equation, we see that 1 mole of Mg reacts with 2 moles of H+ to produce 1 mole of H2.
- Since the amount of metal is not given, we need to convert the given mass of metal to moles.
- The molar mass of Mg is approximately 24.31 g/mol.
- Using the given mass of metal (0.052 g) and the molar mass of Mg, we can calculate the moles of Mg:
Moles of Mg = (0.052 g) / (24.31 g/mol) = 0.002136 mole
- Since the reaction is 1:2 between Mg and H2, the theoretical moles of H2 produced is twice the moles of Mg:
Theoretical moles of H2 = 2 * Moles of Mg = 2 * 0.002136 mole = 0.004272 mole
2. Calculate the percent yield:
- Percent yield is calculated by dividing the actual yield (0.001179 mole) by the theoretical yield (0.004272 mole) and multiplying by 100%:
Percent yield = (Actual yield / Theoretical yield) * 100%
Percent yield = (0.001179 mole / 0.004272 mole) * 100%
Percent yield ≈ 27.58%
Therefore, the average percent yield of H2 in the reaction is approximately 27.58%.