Question #b7940

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Question #b7940

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Answer 1 · 2015-01-22T10:39:30

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 " $mg$ $"mg"$ " in your question stands for. If indeed you are dealing with $100 mL$ $"100 mL"$ of a $1.0 M$ $"1.0 M"$ $HCl$ $"HCl"$ solution, the number of $HCl$ $"HCl"$ moles you'll have is

$n_{HCl} = C \cdot V = 1.0 \frac{mol}{L} \cdot 100 \cdot 10^{- 3} L = 0.10$ $n_("HCl") = C * V = "1.0" "mol"/"L" * 100 * 10^(-3) "L" = 0.10$ $moles$ $"moles"$

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

$M g_{(s)} + 2 H C l_{(a q)} \to M g C l_{2 (a q)} + H_{2 (g)}$ $Mg_((s)) + 2HCl_((aq)) -> MgCl_(2(aq)) + H_(2(g))$

Notice that you have a $1 : 1$ $1:1$ mole ratio between $Mg$ $"Mg"$ and $H_{2}$ $"H"_2$ , which means that every mole of the former will produce 1 mole of the latter. Likewise, keep an eye on the mole ratio between $Mg$ $"Mg"$ and $HCl$ $"HCl"$ , in case one of them turns out to be a limiting reagent.

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

$1.2 g \cdot \frac{1 mole}{24.3 g} = 0.0494$ $"1.2 g" * ("1 mole")/("24.3 g") = 0.0494$ $moles$ $"moles"$

In this case, the solid will be the limiting reagent by a very small margin, which means that the number of $H_{2}$ $"H"_2$ moles produced will be $0.0494$ $"0.0494"$ .

When you add $0.6 g$ $"0.6 g"$ of $Mg$ $"Mg"$ , you'll of course have half the number of moles, $0.0247$ $"0.0247"$ . This time, the number of $H_{2}$ $"H"_2$ moles produced will be equal to $0.0247$ $"0.0247"$ .

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

$\frac{V}{n} = constant$ $"V"/"n" = "constant"$ , or, in this case, $\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$ $"V"_1/"n"_1 = "V"_2/"n"_2$

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

$\frac{V_{1}}{0.0494} = \frac{V_{2}}{0.0247} \Rightarrow V_{1} = \frac{0.0494}{0.0247} \cdot V_{2}$ $"V"_1/0.0494 = "V"_2/0.0247 => V_1 = 0.0494/0.0247 * V_2$ , or

$V_{1} = 2 \cdot V_{2}$ $V_1 = 2 * V_2$ - two times as many moles = twice the volume.

You could actually calculate $V_{1}$ $"V"_1$ and $V_{2}$ $"V"_2$ , but I think this is sufficient to demonstrate the point.

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