Question

What volume, IN MILLILITERS, of 5.0M HCl solution is required to react with 3.00 g or magnesium metal?

Answer 1

To find the volume of 5.0M HCl solution required to react with 3.00g of magnesium metal, we need to use stoichiometry. The stoichiometric ratio between magnesium and HCl is 1:2. Using the given mass of magnesium and the concentration of HCl solution, we can calculate the volume of HCl solution required.

Step-by-step explanation:

To find the volume of 5.0M HCl solution required to react with 3.00g of magnesium metal, we need to use stoichiometry.

The balanced chemical equation for the reaction between magnesium and HCl is:

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

From the equation, we can see that the stoichiometric ratio between magnesium and HCl is 1:2. This means that it takes 2 moles of HCl to react with 1 mole of magnesium.

To calculate the volume of HCl solution, we need to convert the mass of magnesium to moles.
First, calculate the number of moles of magnesium using its molar mass:

Moles of magnesium = Mass of magnesium / Molar mass of magnesium

Moles of magnesium = 3.00g / 24.31g/mol = 0.123 moles

Now, since the stoichiometric ratio is 1:2, we need twice the moles of HCl to react with the given moles of magnesium
Moles of HCl = 2 * Moles of magnesium = 2 * 0.123 moles = 0.246 moles

Finally, we can use the concentration of the HCl solution to calculate the volume using the formula:Volume of HCl solution = Moles of HCl / Concentration of HCl solution

Volume of HCl solution = 0.246 moles / 5.0M = 0.0492 L = 49.2 mL

Answer 2

The volume of HCl = 2.10⁴ mL

Further explanation

Given

5.0 M(mol/L) HCl solution

3 g of Mg

Required

Volume of HCl

Solution

Reaction

Mg + 2HCl ⇒ MgCl₂ + H₂

mol of Mg (Ar=24 g/mol) :

mol = mass : Ar

mol = 3 : 24

mol = 0.125

From the equation, mol ratio of Mg : HCl = 1 : 2, so mol HCl :

= 2/1 x mol Mg

= 2 x 0.125

= 0.25

Molarity : mol solute per liters solution

M = m x V

V = M : m

V = 5 mol/L : 0.25

V = 20 L = 2.10⁴ mL