If you're dealing with hydrochloric acid and balloons, you must be studying a reaction that produces hydrogen gas. Usually, zinc or magnesium will be used to react with the hydrochloric acid to produce hydrogen gas, which is then collected in a balloon.

I'll show you how to do this using magnesium, since I assume that is what the "#"mg"#" in your question stands for. If indeed you are dealing with #"100 mL"# of a #"1.0 M"# #"HCl"# solution, the number of #"HCl"# moles you'll have is

#n_("HCl") = C * V = "1.0" "mol"/"L" * 100 * 10^(-3) "L" = 0.10# #"moles"#

The balanced equation for the chemical reaction looks like this (I won't go into the net ionic equation):

#Mg_((s)) + 2HCl_((aq)) -> MgCl_(2(aq)) + H_(2(g))#

Notice that you have a #1:1# mole ratio between #"Mg"# and #"H"_2#, which means that every mole of the former will produce 1 mole of the latter. Likewise, keep an eye on the mole ratio between #"Mg"# and #"HCl"#, in case one of them turns out to be a limiting reagent.

So, you start by adding #"1.2 g"# of #"Mg"# to the solution. The number of moles of #"Mg"# you've added is

#"1.2 g" * ("1 mole")/("24.3 g") = 0.0494# #"moles"#

In this case, the solid will be the limiting reagent by a very small margin, which means that the number of #"H"_2# moles produced will be #"0.0494"#.

When you add #"0.6 g"# of #"Mg"#, you'll of course have half the number of moles, #"0.0247"#. This time, the number of #"H"_2# moles produced will be equal to #"0.0247"#.

Here's where Avogadro's law becomes very useful. You know that at **constant pressure** and **constant temperature**, the volume an ideal gas occupies is directly proportional to the number of moles of that gas present. Mathematically, this is expressed as

#"V"/"n" = "constant"#, or, in this case, #"V"_1/"n"_1 = "V"_2/"n"_2#

If you add the values for the moles obtained, you'll get

#"V"_1/0.0494 = "V"_2/0.0247 => V_1 = 0.0494/0.0247 * V_2#, or

#V_1 = 2 * V_2# - *two times as many moles = twice the volume*.

You could actually calculate #"V"_1# and #"V"_2#, but I think this is sufficient to demonstrate the point.