Question

If 2.40 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

Answer:

$-112.876J$

Explanation:

In order to solve this question, we would need to incorporate Stoichiometry, which involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Here's a balanced equation for the reaction:

$16KNO_3(s) + 24C(s) + S_8(s) \to 24CO_2(g) + 8N_2(g) + 8K_2S(s)$

Let us define $P - V$ work as;

$w_{pv} = - P_{external} \triangle Volume$

where  $\triangle (Volume) = (V_{final} - V_{initial})$

External pressure is given as  $1.00$$atm$ , therefore the work solely depends on the change in volume and since the reactants are solids, none of the reactants contribute to the volume. Hence,  $V_i = 0.$

To find the volume of the products, we need to first find the amount of moles of the product made from  $2.40_gKNO_3,$ using the molar mass of  $KNO_3$  which is 101.1032 g/mol  

$2.40_gKNO_3 . {\frac{1molKNO_3}{101.1032_g}} = 0.0237molKNO_3$

Now let us convert moles of  $KNO_3$  into moles of $CO_2$ and $N_2$  using the stoichiometric ratios from our balanced equation of the reaction.

$0.0237molKNO_3 . {\frac{24molCO_2}{16molKNO_3}} = 0.0356molCO_2$

$0.0237molKNO_3 . {\frac{8molN_2}{16molKNO_3}} = 0.01185molN_2$

$K_2S$ is not factored into the volume calculation because it is a solid.

Now let us also convert the moles of  $CO_2$  and  $N_2$ into grams using their respective molar masses.

$0.0356molCO_2 . {\frac{44.01_g}{1molCO_2}} = 1.567_gCO_2$

$0.01185molN_2 . {\frac{28.014_g}{1molN_2}} = 0.332_gN_2$

We will now proceed to convert grams into volume using the density values provided.

$1.567_gCO_2 . {\frac{1L}{1.830_g}} = 0.856LCO_2$

$0.332_gN_2 . {\frac{1L}{1.165_g}} = 0.285LN_2$

Summing up the two volumes, we get the final volume

$0.856L + 0.258L = 1.114L = V_f$

Plugging everything into the $w_{pv}$ equation, we get:

$w_{pv} = -1atm(1.114L - 0L) = -1.114L.atm$

Finally, let us convert $L.atm$ into joules using the conversion rate of;

$1L.atm = 101.325J\\-1.114L.atm. {\frac{101.325J}{1L.atm}} = -112.876J$

Answer 2

The amount of P-V work that the gases would do against the given external pressure is -116.64 Joules.

Given the following data:

• Mass of $KNO_3$ = 2.40 grams.

• External pressure = 1.00 atm.

• Density of nitrogen = 1.165 g/L.

• Density of carbon dioxide = 1.830 g/L.

• Temperature = 20°C.

Scientific data:

• Molar mass of $KNO_3$ = 101.1032 g/mol.

• Molar mass of $N_2$ = 28.01 g/mol.

• Molar mass of $C0_2$ = 44.01 g/mol.

To calculate the amount of P-V work that the gases would do against the given external pressure:

The balanced equation for this chemical reaction.

In this scenario, the properly balanced equation for the chemical reaction of $KNO_3$ with sulfur ($S_8$) and carbon (C) is:

                  $KNO_3+S_8 +24C \rightarrow 24C0_2 + 8N_2 +K_2S$

Next, we would calculate the number of moles of $KNO_3$ that was used during this chemical reaction;

$Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{2.40}{101.1032}$

Number of moles = 0.0237 moles.

By stoichiometry:

0.0237 moles of $KNO_3$  = $\frac{24\; of \;CO_2}{16\;of \;K N0_3} \times 0.0237$ = 0.0356 moles of $CO_2$.

0.0237 moles of $KNO_3$  = $\frac{16\; of \;N_2}{8\;of \;K N0_3} \times 0.0237$ = 0.0119 moles of $N_2$.

Next, we calculate the mass of nitrogen and carbon dioxide gases in grams:

For carbon dioxide:

$Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 0.0356 \times 44.01$

Mass = 1.567 grams.

For nitrogen:

$Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 0.0119 \times 28.01$

Mass = 0.333 grams.

Next, we would determine the volume of nitrogen and carbon dioxide gases in liters:

For carbon dioxide:

$Volume = \frac{Mass}{Density} \\\\Volume = \frac{1.567}{1.830}$

Volume = 0.8563 liters.

For nitrogen:

$Volume = \frac{Mass}{Density} \\\\Volume = \frac{0.333}{1.165}$

Volume = 0.2858 liters.

For the final volume:

$Final\;volume = 0.8653+0.2858$

Final volume = 1.1511 Liters.

How to calculate the P-V work by a gas.

Mathematically, the work done by a gas is given by the formula:

$Work = -P \Delta V$

Where:

• P is the pressure.

• $\Delta V$ is the change in volume.

Substituting the given parameters into the formula, we have;

$Work = -1 \times 1.1511$

Work = -1.1511 Latm.

Note: 1 Latm = 101.325 Joules.

Work = $-1.1511 \times 101.325$

Work = -116.64 Joules.

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