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Feliz [49]
3 years ago
14

For this reaction, C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O, the ∆H is –2200 kJ. If two moles of C3H8(g) reacted with excess oxygen,

what would be true?
A) 4400 kj of heat released into surroundings
B) 4400 kj of heat absorbed by system
C) 1100 kJ heat released into surroundings
D) 1100 kJ heat absorbed by system
Chemistry
1 answer:
BartSMP [9]3 years ago
7 0

A) 4400 kj of heat released into surroundings

<h3>Further explanation</h3>

Reaction

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O, the ∆H is –2200 kJ

Reaction exothermic( ∆H=-, released heat to surrounding) and for combustion of 1 mole of C3H8

So for two moles of C3H8, the enthalpy :

\tt -2200\times 2=-4400~kJ

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D

Explanation:

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. The compound Fe(NO3)3(s) is classified as a(n):?. a. polyatomic compound. b. molecular compound. c. ionic compound. d. multi-a
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5 0
3 years ago
Automobile airbags contain solid sodium azide, NaN3, that reacts to produce nitrogen gas when heated, thus inflating the bag. 2N
Vitek1552 [10]

Answer : The value of work done for the system is 1144.69 J

Explanation :

First we have to calculate the moles of NaN_3

\text{Moles of }NaN_3=\frac{\text{Mass of }NaN_3}{\text{Molar mass of }NaN_3}

Molar mass of NaN_3 = 65.01 g/mole

\text{Moles of }NaN_3=\frac{20.2g}{65.01g/mole}=0.311mole

Now we have to calculate the moles of nitrogen gas.

The balanced chemical reaction is,

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

From the balanced reaction we conclude that

As, 2 mole of NaN_3 react to give 3 mole of N_2

So, 0.311 moles of NaN_3 react to give \frac{0.311}{2}\times 3=0.466 moles of N_2

Now we have to calculate the volume of nitrogen gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of N_2 gas = 1.00 atm

V = Volume of N_2 gas = ?

n = number of moles N_2 = 0.466 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 22^oC=273+22=295K

Putting values in above equation, we get:

1.00atm\times V=0.466mole\times (0.0821L.atm/mol.K)\times 295K

V=11.3L

As initially no nitrogen was present. So,

Volume expanded = Volume of nitrogen evolved

Thus,

Expansion work = Pressure × Volume

Expansion work = 1.00 atm × 11.3 L

Expansion work = 11.3 L.atm

Conversion used : (1 L.atm = 101.3 J)

Expansion work = 11.3 × 101.3 = 1144.69 J

Therefore, the value of work done for the system is 1144.69 J

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Answer:

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