# 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?

##
#Fe(s) + 2HCl(aq) -> FeCl_2(aq) + H_2(g)# ;

#DeltaH = -"87.9 kJ"#

#R = 0.0821 (L * atm)/(K* mol)#

##### 1 Answer

#### Explanation:

The idea here is that you need to use the enthalpy change of reaction, **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

#"Fe"_ ((s)) + 2"HCl"_ ((aq)) -> "FeCl"_ (2(aq)) + "H"_ (2(g)) uarr" "DeltaH = -"87.9 kJ"#

This tells you that when **mole** of iron reacts with **moles** of hydrochloric acid, **mole** of hydrogen gas is produced and **given off**, hence the *minus sign* used for the value of

In your case, you know that **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

#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

**one mole**occupies

#color(blue)(|bar(ul(color(white)(a/a)"1 mole" = "22.7 L"color(white)(a/a)|))) -># themolar volume of a gas at STP

Use this as a *conversion factor* to calculate the volume occupied by **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, *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

#PV = nRT#

#V/n = (RT)/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

**SIDE NOTE** *If the molar volume of a gas at STP is given to you as* *which is the value that corresponds to the old definition of STP conditions, i.e. a pressure of*

*and a temperature of*

*simply redo the final calculation using*

*instead of*