How many milliliters of a 2.5 hydrochloric acid solution would be needed to react completely with 32.0 grams of magnesium metal

Here is a worked example of a stoichiometry problem. Remember M x L = moles.

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To determine the number of milliliters of a 2.5 hydrochloric acid (HCl) solution required to react completely with 32.0 grams of magnesium (Mg) metal, we need to use stoichiometry.

1. First, let's write the balanced chemical equation for the reaction between hydrochloric acid and magnesium:
Mg + 2HCl -> MgCl2 + H2

2. Calculate the molar mass of magnesium (Mg) and hydrochloric acid (HCl):
Mg: 24.31 g/mol
HCl: 36.46 g/mol (1 hydrogen + 1 chlorine)

3. Determine the number of moles of magnesium (Mg) from the given mass:
Moles of Mg = mass of Mg / molar mass of Mg
= 32.0 g / 24.31 g/mol
≈ 1.316 moles

4. According to the balanced equation, one mole of magnesium reacts with two moles of hydrochloric acid. Therefore, the number of moles of HCl required is twice the number of moles of Mg:
Moles of HCl = 2 × Moles of Mg
= 2 × 1.316 moles
≈ 2.632 moles

5. Now, we need to find the volume of the HCl solution in milliliters. To do this, we need to know the concentration of the HCl solution. If the concentration is given as 2.5 M (2.5 moles per liter), we can proceed as follows:
Moles of HCl = Concentration of HCl × Volume of HCl solution in liters

Rearranging the equation to solve for the volume of HCl solution:
Volume of HCl solution in liters = Moles of HCl / Concentration of HCl

6. Calculate the volume of HCl solution in liters:
Volume of HCl solution in liters = 2.632 moles / 2.5 M
≈ 1.053 liters

7. Finally, convert the volume from liters to milliliters:
Volume of HCl solution in milliliters = Volume in liters × 1000
≈ 1.053 liters × 1000
≈ 1053 milliliters

Therefore, approximately 1053 milliliters of a 2.5 hydrochloric acid solution would be needed to react completely with 32.0 grams of magnesium metal.