In Flask A, you reacted 5.10 g Mg with 0.447 mol HCl. In Flask B, you reacted 24.21 g Mg with 0.998 mol HCl. Which ballon will inflate the most?

These are two limiting reagent problems rolled into one. If you don't know how to do limiting reagent problems here is a link that will show you how to do them step by step. Post your work if you get stuck.

http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html

To determine which balloon will inflate the most, we need to calculate the amount of hydrogen gas produced in each reaction. The reaction between magnesium (Mg) and hydrochloric acid (HCl) produces hydrogen gas (H2) according to the following balanced chemical equation:

Mg + 2HCl → MgCl2 + H2

First, we need to calculate the amount of hydrogen gas produced in each flask using stoichiometry.

In Flask A:
- Mass of Mg = 5.10 g
- Moles of HCl = 0.447 mol

We need to find the limiting reactant, which is the reactant that is completely consumed and determines the amount of product formed. To do this, we compare the moles of Mg and moles of HCl.

The reaction ratio from the balanced equation is 1 mol Mg : 2 mol HCl.
So, for every 1 mole of Mg, we need 2 moles of HCl.

Using the given moles of HCl and the reaction ratio, we can calculate the moles of Mg needed:
Moles of Mg = moles of HCl / reaction ratio
Moles of Mg = 0.447 mol / (2 mol HCl / 1 mol Mg) = 0.2235 mol Mg

Since the moles of Mg present (0.2235 mol) is less than the moles of Mg used (0.447 mol), it is the limiting reactant.

Now, we can calculate the moles of hydrogen gas produced based on the limiting reactant:
Moles of H2 = Moles of Limiting Reactant = 0.2235 mol

In Flask B:
- Mass of Mg = 24.21 g
- Moles of HCl = 0.998 mol

Following the same process as above, we find that the moles of hydrogen gas produced in Flask B is also 0.2235 mol.

Therefore, the amount of hydrogen gas produced in both flasks is equal. As a result, both balloons will inflate the same amount.