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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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fenix001 [56]
Step 1: Write the unbalanced equation,

                                 C₂H₆  +  O₂    →    CO₂  +  H₂<span>O

There are 2 C at left hand side and 1 carbon at right hand side. So, multiply CO</span>₂ by 2 to balance C atoms at both side. So,

                                 C₂H₆  +  O₂    →   2 CO₂  +  H₂O

Now, count number of H atoms at both sides. There are 6 H atoms at left hand side and 2 at right hand side. Multiply H₂O by 3 to balance H atoms.


                                 C₂H₆  +  O₂    →   2 CO₂  +  3 H₂O

At last, balance O atoms. There are 2 O atoms at left hand side and 3 O atoms at right hand side. Multiply O₂ with 1.5 (i.e. 3/2) to balance O atoms. i.e.

                                 C₂H₆  +  3/2 O₂    →   2 CO₂  +  3 H₂O

Hence, the equation is balanced. If you want to make equation fraction free then multiply all equation with 2. i.e.

                           ( C₂H₆  +  3/2 O₂    →   2 CO₂  +  3 H₂O ) × 2

                                2 C₂H₆  +  3 O₂    →   4 CO₂  +  6 H₂O
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