Question

a 1750 ml sample of 2.75 m hcl solution is treated with 4.27 g of magnesium. calculate the concentration of the acid solution after all the metal has reacted. assume that the volume remains unchanged.

Answer 1

After reacting 4.27 g of magnesium with 1750 mL of a 2.75 M HCl solution and assuming the volume remains the same, the concentration of the acid solution decreases to approximately 2.55 M.

Step-by-step explanation:

The question asks us to calculate the concentration of HCl solution after it reacts with magnesium (Mg). We start by figuring out how many moles of Mg reacted, use the stoichiometry of the reaction to find out the moles of HCl consumed, and then calculate the new concentration of HCl.

We start with the mass of Mg given:

• Mass of Mg = 4.27 g

The molar mass of Mg is approximately 24.305 g/mol, so we calculate the moles of Mg:

Moles of Mg = mass of Mg / molar mass of Mg

Moles of Mg = 4.27 g / 24.305 g/mol ≈ 0.1757 mol

The balanced equation for the reaction is:

Mg + 2HCl → MgCl2 + H2

From the equation, 1 mole of Mg reacts with 2 moles of HCl. So the moles of HCl reacted are:

Moles of HCl reacted = 2 * Moles of Mg

Moles of HCl reacted = 2 * 0.1757 mol ≈ 0.3514 mol

The original moles of HCl in the solution can be calculated from its molarity (M) and volume (V):

M1V1 = moles of HCl (initial)

2.75 M * 1.750 L = 4.8125 mol

Now we can find the moles of HCl remaining after the reaction:

Moles of HCl remaining = initial moles of HCl - moles of HCl reacted

Moles of HCl remaining = 4.8125 mol - 0.3514 mol ≈ 4.4611 mol

Assuming the volume hasn't changed, the new concentration (M2) of the HCl solution is:

M2 = moles of HCl remaining / volume of solution

M2 = 4.4611 mol / 1.750 L ≈ 2.55 M

Therefore, the concentration of the acid solution after all the magnesium has reacted is approximately 2.55 M.