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Eva8 [605]
3 years ago
10

Ans with solution...

Chemistry
1 answer:
Licemer1 [7]3 years ago
3 0

Answer:

The answer to your question is:

Vol of NO2 = 11.19 L

Vol of O2 = 2.8 L

Explanation:

Data

N2O5 = 56 g

STP     T = 0°C = 273°K

           P = 1 atm

MW N2O5 = 216 g

Gases law = PV = nRT

Process

                   216 g of N2O5 ---------------- 1 mol

                     54 g               -----------------  x

                    x = (54 x 1) / 216

                    x = 0.25 mol of N2O5

                   2 mol of N2O5 -----------------  4 mol of NO2

                   0.25 mol          ------------------    x

                   x = (0.25 x 4) / 2 = 0.5 mol of NO2

                   V = nRT/P

                   V = (0.5)(0.082)(273) / 1 = 11.19 L

                   2 mol of N2O5 ----------------- 1 O2

                   0.25 N2O5 ----------------------  x

                   x = (0.25 x 1) / 2 = 0.125 mol

                   Vol = (0.125)((0.082)(273) / 1 = 2.8 L

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<u>Explanation:</u>

The ideal gas equation is given as:

.......(1)

where

P = pressure of the gas = 675 torr

V = volume of gas = ?

n = number of moles of gas = 6.45 moles

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Putting values in equation 1, we get:

675torr\times V=6.45mol\times 62.36L.torr/mol.K\times 300K\\\\V=\frac{6.45\times 62.36\times 300}{675}=178.76 L

Hence, the volume of the gas is 178.76 L

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