Question

Based on the following reaction, calculate the volume (in L) of O_3 needed to produce 7.75 g KO_3 Pressure = 110 torr, Temperater- 31.0degreeC The work done to compress a gas is 75 J 25 J of heat is absorbed fron the surroundings what is the energy of the gas in joules? g sample of nickel at 92.8degreeC was mixed with 119.0 g of water at 24.5degreeC in a coffee p calorimeter. The final temperature of the mixture was 28.0degreeC. The specific heat capacity of water is 4.18 J/gdegreeC. Calculate the specific heat capacity of nickel, (assume the I capacity of the calorimeter is zero)

Answer 1

4) As we know the ideal gas equation as

$PV=nRT$

n= number of moles =$\frac{w}{MW}$

w= given mass

MW= molar mass

R= gas constant

T = temperature in Kelvin

Now here we have the reaction as

$5O_3_{(g)}+2KOH\rightarrow 2KO_3+5O_2_{(g)}+H_2O_{(l)}$

Here 2 mole of $KO_3_{(s)}$ are produced from 5 mole of $O_3_{(g)}$

1 mole = molar mass in gm

2 mole of $KO_3_{(s)}$ = 2 x 87.1 = 174.2 gm

5 mole of $O_3_{(g)}$ = 5 x 48 = 240 gm

That means 174.2 gm of $KO_3_{(s)}$ is produced from 240 gm of $O_3_{(g)}$

then 7.75 gm of $KO_3_{(s)}$ will be produced from the amount of $O_3_{(g)}$ as

$\\=\frac{240}{174.2}\times 7.75\\ =10.67gm$

Now the moles of $O_3_{(g)}$ found as

$\\=\frac{w}{MW}\\ =\frac{10.67}{48}\\ =0.22 mol$

Again we know

$R=62.36LTorrK^{-1}Mol^{-1}$

P = 110 torr

n = 0.22 moles

T = 31 ⁰ C = 31+273 =304 K

Now apply

$PV=nRT$

$\\110torr\times V=0.22mol\times 62.36 LtorrK^{-1}mol^{-1}\times 304K\\ 110\times V=4170 L\\ V=\frac{4170.63}{110}L\\ V=37.91 L$

Hence the volume of $O_3_{(g)}$ is 37.91 L

Answer.