# The reaction of iron with hydrochloric acid is represented by the following thermochemical equation. If, in a particular experiment, 7.36 kJ of heat was released at constant pressure, what volume of H_2(g), measured at STP, was produced?

## $F e \left(s\right) + 2 H C l \left(a q\right) \to F e C {l}_{2} \left(a q\right) + {H}_{2} \left(g\right)$; $\Delta H = - \text{87.9 kJ}$ $R = 0.0821 \frac{L \cdot a t m}{K \cdot m o l}$

##### 1 Answer
Jul 12, 2016

$\text{1.90 L}$

#### Explanation:

The idea here is that you need to use the enthalpy change of reaction, $\Delta H$, to figure out the number of moles of iron, and consequently of hydrochloric acid, that took part in the reaction.

This will then allow you to figure out how many moles of hydrogen gas were produced by the reaction. At that point, you can use the molar volume of a gas at STP to calculate the volume of the hydrogen gas.

So, the thermochemical equation that describes this single replacement reaction looks like this

$\text{Fe"_ ((s)) + 2"HCl"_ ((aq)) -> "FeCl"_ (2(aq)) + "H"_ (2(g)) uarr" "DeltaH = -"87.9 kJ}$

This tells you that when $1$ mole of iron reacts with $2$ moles of hydrochloric acid, $1$ mole of hydrogen gas is produced and $\text{87.9 kJ}$ of heat are given off, hence the minus sign used for the value of $\Delta H$.

In your case, you know that $\text{7.36 kJ}$ of heat were given off by the reaction. This should automatically tell you that the reaction consumed less than one mole of iron.

More specifically, it consumed

7.36 color(red)(cancel(color(black)("kJ given off"))) * "1 mole Fe"/(87.9 color(red)(cancel(color(black)("kJ given off")))) = "0.08373 moles Fe"

You know how many moles of iron were consumed in the reaction, so use the $1 : 1$ mole ratio that exists between iron and hydrogen gas to say that the reaction produced

0.08373 color(red)(cancel(color(black)("moles Fe"))) * "1 mole H"_2/(1color(red)(cancel(color(black)("mole Fe")))) = "0.08373 moles H"_2

Now, STP conditions are currently defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$. The thing to remember here is that when an ideal gas is kept under STP conditions, one mole occupies $\text{22.7 L}$.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mole" = "22.7 L} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ the molar volume of a gas at STP

Use this as a conversion factor to calculate the volume occupied by $0.08373$ moles of hydrogen gas kept under STP conditions

0.08373 color(red)(cancel(color(black)("moles H"_2))) * "22.7 L"/(1color(red)(cancel(color(black)("mole H"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("1.90 L")color(white)(a/a)|)))

The answer is rounded to three sig figs.

Notice that the problem provides you with the value of the universal gas constant, $R$. You don't really need to use it here because you can use the molar volume of gas at STP.

In fact, you can prove that the molar volume of a gas at STP is $\text{22.7 L}$ by using the ideal gas law equation

$P V = n R T$

$\frac{V}{n} = \frac{R T}{P}$

Plug in your values to find

V/n = (0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm")))) = "22.7 L mol"^(-1)

This is why one mole of an ideal gas occupies $\text{22.7 L}$ at STP.

SIDE NOTE If the molar volume of a gas at STP is given to you as $\text{22.4 L}$, which is the value that corresponds to the old definition of STP conditions, i.e. a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$, simply redo the final calculation using $\text{22.4 L}$ instead of $\text{22.7 L}$.