Answer:2.50 moles of magnesium will consume approximately 182.30 grams of hydrochloric acid in the given reaction.
Explanation:To find out how many grams of hydrochloric acid (HCl) are consumed when 2.50 moles of magnesium (Mg) react with it, you can use stoichiometry and the balanced chemical equation:
Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)
From the balanced equation, you can see that 1 mole of magnesium (Mg) reacts with 2 moles of hydrochloric acid (HCl).
Now, let's use this information to calculate the moles of HCl required to react with 2.50 moles of Mg:
Moles of HCl = (2.50 moles Mg) * (2 moles HCl / 1 mole Mg)
Moles of HCl = 2.50 moles * 2
Moles of HCl = 5.00 moles
Now that we know we need 5.00 moles of HCl, we can calculate the grams of HCl needed using the molar mass of HCl:
The molar mass of HCl is the sum of the atomic masses of hydrogen (H) and chlorine (Cl):
Molar mass of HCl = 1.01 g/mol (for hydrogen) + 35.45 g/mol (for chlorine)
Molar mass of HCl = 36.46 g/mol
Now, calculate the grams of HCl:
Grams of HCl = (5.00 moles) * (36.46 g/mol)
Grams of HCl = 182.30 grams
So, 2.50 moles of magnesium will consume approximately 182.30 grams of hydrochloric acid in the given reaction.