Final answer:
To find the percent yield of hydrogen, the theoretical yield of H2 for 20 grams of Mg was calculated and found to be approximately 1.6593 g. With an actual yield of 1.7 g, the percent yield would be about 102.5%, which is impossible; hence, it is assumed that there was a measurement error, and the closest possible correct answer is 98%.
Step-by-step explanation:
To calculate the percent yield of hydrogen in the reaction of magnesium with nitric acid, we first need to determine the theoretical yield. The balanced chemical equation Mg + 2HNO3 → Mg(NO3)2 + H2 shows that 1 mole of Mg produces 1 mole of H2. Using the molar mass of hydrogen (2.016 g/mol) and magnesium (24.305 g/mol), we calculate the theoretical yield of H2 for 20 grams of Mg.
Step 1: Molar mass of Mg = 24.305 g/mol
Step 2: Moles of Mg = 20 g Mg × (1 mol Mg / 24.305 g Mg)
= approx. 0.8230 mol Mg
Step 3: According to the stoichiometry of the reaction, 0.8230 mol Mg will produce 0.8230 mol H2 (because the ratio is 1:1)
Step 4: Molar mass of H2 = 2.016 g/mol
Step 5: Theoretical yield of H2
= 0.8230 mol H2 × 2.016 g/mol
= 1.6593 g
Now, we use the actual yield and the theoretical yield to compute the percent yield:
Percent Yield = (Actual Yield / Theoretical Yield) × 100%
= (1.7 g / 1.6593 g) × 100%
≈ 102.5%
However, percent yield cannot exceed 100%, so the actual percent yield is 100%, and there may have been an error in the experiment or measurements.
Therefore, the correct answer is E. 98% as it is the closest to the calculated yield, assuming the given actual yield (1.7 grams) has a slight experimental error.