The amount of Mg(OH)2 present after the reaction is complete is 0.136 moles of Mg(OH)2.
The equation of the reaction is;
2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)
Number of moles of Mg(OH)2 = 8.00 g/58 g/mol = 0.138 moles
Number of moles of HNO3 = 0.205 M × 24.2 mL/1000 = 0.00496 moles
Given that;
2 moles of HNO3 reacts with 1 mole of Mg(OH)2
0.00496 moles of HNO3 reacts with 0.00496 moles × 1 mole /2 moles = 0.00248 moles of Mg(OH)2
Hence, Mg(OH)2 is the reactant in excess.
The amount of Mg(OH)2 remaining = Amount present - Amount reacted
Hence; 0.138 moles - 0.00248 moles = 0.136 moles of Mg(OH)2
Learn more: https://brainly.com/question/9743981
