Question

A 2.50 x 10^3 sample of 2.25M HCI solution is treated with 4.30 g of magnisium. calclate the concentration of the acid solution aftrer all the metal is reacted. Assume the volume remains unchanged.

Answer 1

Solution:- Assuming the given sample volume of HCL is in "mL" unit, i.e. (2.50 x 10³ mL) or we can say 2.50 L.

Now, the equation of the reaction:

2HCl(l) + Mg(s) -----> MgCl₂(aq) + H₂(g)

Now find the limiting reactant:

HCl = 2.50(L) x 2.25 (mol/L) = 5.625 mol.

Mg = 4.30 (g) / (24.305 g/mol) = 0.1769 mol.

Here, Mg is the limiting reactant.

0.1769 mol of Mg x (1 mol MgCl₂) / (1 mol of Mg) = 0.1769 mol MgCl₂

0.1769 mol MgCl₂ x (2 mol HCl) / (1 mol MgCl₂) = 0.3538 mol HCl

Remaining moles of HCl = 1 - 0.3538 = 0.6462 mol HCl.

So, concentration of the acid solution = 0.6462 mol / 2.50 L = 0.2584 M

Concentration of acid solution after metal is reacted = 0.258 M