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?

Feb 24, 2018

${w}_{P V} = - 166.173 J$

Explanation:

To solve this question, you need to incorporate stoichiometry. The first step is writing out the whole balanced reaction. Given $K N {O}_{3}$, ${S}_{8}$, and $C$, the reaction will be:

$16 K N {O}_{3} \left(s\right) + 24 C \left(s\right) + {S}_{8} \left(s\right) \rightarrow 24 C {O}_{2} \left(g\right) + 8 {N}_{2} \left(g\right) + 8 {K}_{2} S \left(s\right)$

Setting this equation up is annoying, but when you realize that the other product is ${K}_{2} S$ because $K \mathmr{and} S$ 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 ${w}_{P V} = - {P}_{e x t e r n a l} \Delta V o l u m e$.

Note: $\Delta \left(V o l u m e\right) = \left({V}_{F I N A L} - {V}_{I N I T I A L}\right) \mathmr{and} \left({V}_{f} - {V}_{i}\right)$

External pressure is given as $1.00 a t m$ 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 $3.45 g K N {O}_{3}$.

Convert $K N {O}_{3}$ into moles using its molar mass of $101.102 g$.
($K N {O}_{3}$ is the limiting reactant because there is "sufficient" (i.e. excess) sulfur and carbon.)

$3.45 g K N {O}_{3} \cdot \frac{1 m o l K N {O}_{3}}{101.102 g} = .0341 m o l K N {O}_{3}$

Now convert moles $K N {O}_{3}$ into moles $C {O}_{2}$ 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

$.0341 m o l K N {O}_{3} \cdot \frac{8 m o l {N}_{2}}{16 m o l K N {O}_{3}} = .01705 m o l {N}_{2}$

${K}_{2} S$ 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.

$.05115 m o l C {O}_{2} \cdot \frac{44.01 g}{1 m o l C {O}_{2}} = 2.251 g C {O}_{2}$

$.01705 m o l {N}_{2} \cdot \frac{28.014 g}{1 m o l {N}_{2}} = .478 g {N}_{2}$

Now convert grams into volume using the density values given.

$2.251 g C {O}_{2} \cdot \frac{1 L}{1.830 g} = 1.230 L o f C {O}_{2}$

$.478 g {N}_{2} \cdot \frac{1 L}{1.165 g} = .410 L o f {N}_{2}$

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

$1.230 L + .410 L = 1.640 L = {V}_{f}$

Now plug everything into the ${w}_{P V}$ equation.

${w}_{P V} = - 1 a t m \left(1.640 L - 0 L\right) = - 1.640 L \cdot a t m$

For the last step, convert $L \cdot a t m$ into joules using the conversion $1 L \cdot a t m = 101.325 J$

$- 1.640 L \cdot a t m \cdot \frac{101.325 J}{1 L \cdot a t m} = - 166.173 J$