Question

If 3.45 g of KNO3 reacts with sufficient sulfur (S8) and carbon (C), how much P-V work will the gases do against an external pressure of 1.00 atm given the densities of nitrogen and carbon dioxide are 1.165 g/L and 1.830 g/L, respectively, at 20°C?

Answer 1

`w_(PV)= -166.173 J`

Explanation:

To solve this question, you need to incorporate stoichiometry. The first step is writing out the whole balanced reaction. Given `KNO_3`, `S_8`, and `C`, the reaction will be:

`16KNO_3(s) + 24C(s) + S_8(s) rarr 24CO_2(g) + 8N_2(g) +8K_2S(s)`

Setting this equation up is annoying, but when you realize that the other product is `K_2S` because `K and S` are the only reactants not in the products given, and they occur in a `1:2` ratio due to their oxidation states, balancing becomes a little easier.

The definition of `P-V` work is `w_(PV)= -P_(external)DeltaVolume`.

Note: `Delta(Volume) = (V_(FINAL) - V_(INITIAL)) or (V_f - V_i)`

External pressure is given as `1.00 atm` so the work only depends on the change of volume.

Since the reactants are all solids, none of the reactants contribute to the volume. Hence, `V_i = 0`.

To find the volume of the products, you need to use the reaction to find the amount of moles of the product made from `3.45g KNO_3`.

Convert `KNO_3` into moles using its molar mass of `101.102g`.
(`KNO_3` is the limiting reactant because there is "sufficient" (i.e. excess) sulfur and carbon.)

`3.45gKNO_3 * (1mol KNO_3)/(101.102g)= .0341 mol KNO_3`

Now convert moles `KNO_3` into moles `CO_2` and `N_2` using the stoichiometric ratios from the balanced equation.

#.0341 mol KNO_3 * (24 mol CO_2)/(16 mol KNO_3) = .05115 mol CO_2#

`.0341 mol KNO_3 * (8 mol N_2)/(16 mol KNO_3) = .01705 mol N_2`

`K_2S` doesn't factor into the volume because it is a solid, so you don't need to include it into the volume calculation.

Now convert the moles into grams using molar masses.

`.05115 mol CO_2 * (44.01g)/(1 mol CO_2) = 2.251 g CO_2`

`.01705 mol N_2 * (28.014g)/(1 mol N_2) = .478 g N_2`

Now convert grams into volume using the density values given.

`2.251 g CO_2 * (1 L)/(1.830 g) = 1.230 L of CO_2`

`.478 g N_2 * (1 L)/(1.165g) = .410 L of N_2`

To find the final volume, just sum the two volumes.

`1.230L + .410L = 1.640L = V_f`

Now plug everything into the `w_(PV)` equation.

`w_(PV)= -1atm(1.640L- 0L) = -1.640 L*atm`

For the last step, convert `L*atm` into joules using the conversion `1L*atm= 101.325 J`

`-1.640L*atm *(101.325J)/(1L*atm)= -166.173J`