<h3>
Answer:</h3>
43.27 g Mg
<h3>
Explanation:</h3>
The balanced equation for the reaction between magnesium metal and hydrochloric acid is;
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
From the equation;
1 mole of magnesium reacts with 2 moles of HCl
We are given;
3.56 moles of Mg and 3.56 moles of HCl
Using the mole ratio;
3.56 moles of Mg would react with 7.12 moles of HCl, and
3.56 moles of HCl would react with 1.78 moles of Mg
Therefore;
The amount of magnesium was in excess;
Moles of Mg left = 3.56 moles - 1.78 moles
= 1.78 moles
But; 1 mole of Mg = 24.305 g/mol
Therefore;
Mass of magnesium left = 1.78 moles × 24.305 g/mol
= 43.2629 g
= 43.27 g
Thus, the mass of magnesium that remained after the reaction is 43.27 g