Question

I’m sorry but every time I get this question wrong and I need help

Answer 1

Step 1 - Useful assumptions to solve the question

Note that we do not have an equation that describes properly the behavior of a real gas. In fact, there a lot of them, but all too complicated. So our first useful assumption is that the produced gas, CO, is an ideal gas.

This is useful because the behavior of an ideal gas and its dependence on temperature, volume and pressure can be simply described by the equation bellow:

`pV=\text{nRT }`

Note also that while we have enough information concerning the temperature and the volume, we do not know the pressure. Since we are doing an experiment in non-controlled atmosphere, we can just assume that the pressure is 1 atm, the "ambient pressure" at sea level.

Let's recapitulate:

a) The gas can be considered ideal

b) The experiment is being done at sea level therefore p=1 atm

Step 2 - Using the information available and the assumptions to solve the problem

As we previously saw, the behavior of any ideal gas can be described by:

`pV=\text{nRT}`

From the exercise as well as from our assumptions, we already know that:

`\begin{gathered} T=35\degree C(308\text{ K)} \\ \\ V=58.4\text{ ml = 0.0584 L} \\ \\ n\text{ = ?} \\ \\ p=1\text{ atm} \\ \\ R=0.082\text{ atm.L/K.mol (universal gas constant)} \end{gathered}`

Since we want the mass of gas that was formed, we can substitute the number of moles (n) by the following expression, in which m represents the mass and M the molar mass of CO:

`n=(m)/(M)\rightarrow pV=(mRT)/(M)`

For CO, M=28 g/mol. Substituting all the values on the equation, we get:

`1*0.0584=(m*0.082*308)/(28)`

Working on the math, we obtain thus:

`m=(28*0.0584)/(308*0.082)=(1.63)/(25.25)=6.4*10^(-2)\text{ g of CO}`

The mass of CO that was formed is thus 6.4*10^(-2) g.