# Question 3b58a

##### 1 Answer
Mar 13, 2016

Here's what I got.

#### Explanation:

Your starting point here will be the balanced chemical equation for this single replacement reaction

${\text{Mg"_text((s]) + color(red)(2)"HCl"_text((aq]) -> "MgCl"_text(2(aq]) + "H}}_{\textrm{2 \left(g\right]}} \uparrow$

Notice that you have a $1 : \textcolor{red}{2}$ mole ratio between magnesium metal and hydrochloric acid. This tells you that the reaction will always consume twice as many moles of hydrochloric acid as you have moles of magnesium metal taking part in the reaction.

Now, use magnesium's molar mass to determine how many moles you have in that $\text{2.4-g}$ sample

2.4 color(red)(cancel(color(black)("g"))) * "1 mole Mg"/(24.305color(red)(cancel(color(black)("g")))) = "0.09875 moles Mg"

Next, use the molarity and volume of the hydrochloric acid solution to determine how many moles you're supplying to the reaction.

As you know, molarity tells you how many moles of solute, which in your case is hydrochloric acid, you get per liter of solution. This means that you have

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This will get you

${n}_{H C l} = \text{0.50 mol" color(red)(cancel(color(black)("L"^(-1)))) * 359 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.1795 moles HCl}$

Now, do you have enough moles of hydrochloric acid to ensure that all the moles of magnesium metal react?

Use the aforementioned mole ratio to find out. That many moles of magnesium would require

0.09875color(red)(cancel(color(black)("moles Mg"))) * (color(red)(2)color(white)(a)"moles HCl")/(1color(red)(cancel(color(black)("mole Mg")))) = "0.1975 moles HCl"

However, you only have $0.1795$ moles of hydrochloric acid available. This means that hydrochloric acid will act as a limiting reagent, i.e. it will be completely consumed before all the magnesium metal gets a chance to react.

More specifically, the reaction will only consume

0.1795color(red)(cancel(color(black)("moles HCl"))) * "1 mole Mg"/(color(red)(2)color(red)(cancel(color(black)("moles HCl")))) = "0.08975 moles Mg"

Now, you have a $1 : 1$ mole ratio between magnesium metal and hydrochloric acid, which means that this many moles of magnesium would theoretically produce $0.08975$ moles of hydrogen gas.

This would be the reaction's theoretical yield, i.e. what you get if all the moles of magnesium that take part in the reaction end up forming an equal number of moles of hydrogen gas, i.e. if the reaction has a 100% yield.

Now, you know that the reaction produced $\text{125.0 mL}$ of hydrogen gas at STP. As you know, STP conditions are usually given as a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$.

Under these specific conditions for pressure and temperature, one mole of any ideal gas occupies exactly $\text{22.4 L}$ - this is know as the molar volume of gas at STP.

color(red)(!)SIDE NOTE This value is based on the old definition of STP. The current definition implies a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

Under these conditions, one mole of any ideal gas occupies $\text{22.7 L}$.

Use this to find the number of moles of hydrogen produced by the reaction - use the conversion factor

$\text{1 L" = 10^3"mL}$

to convert the volume of the gas from milliliters to liters.

125.0 * 10^(-3)color(red)(cancel(color(black)("L"))) * overbrace("1 mole H"_2/(22.4color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas at STP")) = "0.005580 moles H"_2

So, the reaction should have produced $0.08975$ moles of hydrogen gas, but it ended up producing only $0.005507$ moles.

The reaction's percent yield is defined as

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{% yield" = "what you actually get"/"what you should theoretically get} \times 100 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in your values to get

"% yield" = (0.005580color(red)(cancel(color(black)("moles"))))/(0.08975color(red)(cancel(color(black)("moles")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"6.2%"color(white)(a/a)|)))#

The answer is rounded to two sig figs.

If you ask me, this value is way too low, so you might want to check the values given to you for the mass of magnesium, molarity and volume of the solution, and volume of gas produced at STP.