Final answer:
To find the minimum volume of 6 M hydrochloric acid needed to react with 0.2000 g of aluminum, calculate the moles of aluminum, determine the moles of HCl required from the stoichiometry of the reaction, and then use the molarity to find the volume. The calculation reveals that 3.705 mL of hydrochloric acid is needed.
Step-by-step explanation:
To calculate the minimum volume of 6 M hydrochloric acid required to completely react with 0.2000 g of pure aluminum, we first need the balanced chemical equation for the reaction between aluminum and hydrochloric acid:
2 Al(s)+6 HCl(aq) →2 AlCl₃(aq)+3 H₂(g)
Next, we'll convert the mass of aluminum to moles by using its molar mass:
- Molar mass of Al = 26.98 g/mol
- Moles of Al = 0.2000 g / 26.98 g/mol = 0.00741 mol
Referring to the balanced equation, 2 moles of Al react with 6 moles of HCl. So, 0.00741 moles of Al will react with 3 times as many moles of HCl:
- Moles of HCl required = 3 × 0.00741 mol = 0.02223 mol
To find the volume of HCl, we'll use the molarity formula:
- Molarity (M) = moles/Liter (L)
- 6 M = 0.02223 mol / V (where V is the volume in liters)
- V = 0.02223 mol / 6 M
- V = 0.003705 L
Therefore, the minimum volume of 6 M hydrochloric acid required to completely react with 0.2000 g of pure aluminum is 0.003705 liters or 3.705 mL.