Study Notes on Magnesium Metal and Hydrochloric Acid Reaction

Magnesium Metal and Hydrochloric Acid Reaction

1. Overview of the Reaction

  • The chemical reaction that occurs when magnesium metal reacts with hydrochloric acid can be expressed with the following balanced equation:
    • Mg (s)+2HCl (aq)→MgCl<em>2(aq)+H</em>2(g)\text{Mg (s)} + 2 \text{HCl (aq)} \rightarrow \text{MgCl}<em>2 (aq) + \text{H}</em>2 (g)
    • This equation demonstrates that one mole of magnesium reacts with two moles of hydrochloric acid to produce one mole of magnesium chloride and one mole of hydrogen gas.

2. Theoretical Yield of Hydrogen Gas

a. Calculation of Theoretical Yield
  • Given:
    • Mass of hydrochloric acid (HCl) = 95.0 g
  • Molar mass of HCl = 36.46 g/mol
  • First, calculate the number of moles of HCl:
    • Moles of HCl=Mass of HClMolar mass of HCl=95.0 g36.46 g/mol≈2.61 mol\text{Moles of HCl} = \frac{\text{Mass of HCl}}{\text{Molar mass of HCl}} = \frac{95.0 \text{ g}}{36.46 \text{ g/mol}} \approx 2.61 \text{ mol}
  • According to the balanced equation, 2 moles of HCl produce 1 mole of H₂ gas. Thus the moles of H₂ that can be produced are:
    • Moles of H2=12×Moles of HCl≈12×2.61extmol≈1.30 mol\text{Moles of H}_2 = \frac{1}{2} \times \text{Moles of HCl} \approx \frac{1}{2} \times 2.61 ext{ mol} \approx 1.30 \text{ mol}
  • Now calculate the mass of hydrogen gas:
    • Molar mass of H₂ = 2.02 g/mol
  • Mass of H<em>2=Moles of H</em>2×Molar mass of H2=1.30 mol×2.02 g/mol≈2.63 g\text{Mass of H}<em>2 = \text{Moles of H}</em>2 \times \text{Molar mass of H}_2 = 1.30 \text{ mol} \times 2.02 \text{ g/mol} \approx 2.63 \text{ g}
  • Theoretical yield of hydrogen gas = 2.63 g

3. Actual Yield of the Reaction

b. Data from Experiment
  • During the experimental setup, the group of students lost some mass due to hydrogen escaping, which resulted in a mass loss of:
    • Mass lost=2.41 g\text{Mass lost} = 2.41 \text{ g}
  • Actual yield is defined as the mass of the product obtained from the reaction, which can be determined as follows:
    • Actual Yield (H2)=Theoretical Yield−Mass lost=2.63 g−2.41 g=0.22 g\text{Actual Yield (H}_2) = \text{Theoretical Yield} - \text{Mass lost} = 2.63 \text{ g} - 2.41 \text{ g} = 0.22 \text{ g}
  • Actual yield of hydrogen gas = 0.22 g

4. Percent Yield of the Experiment

c. Calculation of Percent Yield
  • Percent yield is calculated using the formula:
    • Percent Yield=(Actual YieldTheoretical Yield)×100\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
  • Substituting the known values:
    • Percent Yield=(0.22 g2.63 g)×100≈8.36%\text{Percent Yield} = \left( \frac{0.22 \text{ g}}{2.63 \text{ g}} \right) \times 100 \approx 8.36\%
  • The group's percent yield for their experiment = 8.36%

5. Additional Data from Reaction Between Pitres

  • The document references additional data obtained from another reaction involving pitres. Ensure to gather this data for further examination and analysis.