Question

what volume of 0.750 M hydrochloric acid solution can be prepared from the HCl produced by the reaction of 25.0 g of NaCl with excess sulfuric acid?

NaCl (s) + H2SO4 (L) ---> HCl (g) + NaHSO4 (s)

Answer 1

The balanced reaction of NaCl and H2SO4 is

NaCl) + H2SO4(1) + HCl) + NaHSO4(s)

Amount of NaCl being used = 25.0 g

Molar mass of NaCl = 58.44 g/mol

Hence, number of moles of NaCl being used is

25.09 mass molar mass 0.4278 mol 58.44 g/mol

Note that the other reactant H2SO4 is in excess. Hence, NaCl is the limiting reactant.

Now, from the balanced reaction, 1 mol of NaCl produces 1 mol of HCl. Hence, number of moles of HCl that will form from 0.4278 mol of NaCl is

1 mol HCI x 0.4278 mol NaCl = 0.4278 mol HCI 1 mol Naci

The desired concentration of HCl produced is 0.750 M = 0.750 mol/L.

Hence, the volume of 0.750 M HCl that can be produced from 0.4278 mol of HCl produced is

7 x 0.4278 mol 0.750 mol 0.570 L

Hence, we can prepare 0.570 L of 0.750 M HCl with the produced HCl.