Question

Mg(OH)₂ reacted with dilute hydrochloric acid as following equation. Mg(OH)₂(s) 2HCl(aq) →MgCl₂(aq) 2H₂O(l) calculate the mass of excess reagent remaining at the end of the reaction when 25.14 ml of 0.5 m HCl were reacted with 500 mg of Mg(OH)₂?

Answer 1

To calculate the mass of excess reagent remaining at the end of the reaction, compare the moles of HCl and Mg(OH)₂ to determine the limiting reagent. Then, calculate the mass of excess reagent remaining.

Step-by-step explanation:

To calculate the mass of excess reagent remaining at the end of the reaction, we need to determine the limiting reagent first. The limiting reagent is the reactant that is completely consumed in the reaction, thus determining the maximum amount of product that can be formed. In this case, we need to compare the moles of HCl and Mg(OH)₂ to determine the limiting reagent.

To calculate the moles of HCl, we multiply the volume of the solution by its molarity:

moles of HCl = volume of HCl solution (L) × molarity of HCl (mol/L)

Next, we calculate the moles of Mg(OH)₂ using its molar mass:

moles of Mg(OH)₂ = mass of Mg(OH)₂ (g) / molar mass of Mg(OH)₂ (g/mol)

Finally, we compare the moles of HCl and Mg(OH)₂ to determine the limiting reagent. The reactant that has the smaller number of moles is the limiting reagent, and the other reactant is in excess. From there, we can calculate the mass of excess reagent remaining at the end of the reaction.