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4 years ago

Which of the following statements is true for an exothermic reaction?

Chemistry

1 answer:

Ann [662]4 years ago

3 0

For an exothermic reaction,heat energy is released out.

Enthalpy change = Energy of products - Energy of reactants

Here enthalpy change is negative,as energy of products is less than energy of reactants.

Hence heat is released in exothermic reactions.

And the products have lower potential energy than the reactants, and the enthalpy change is negative.

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A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed

AlekseyPX

(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(a) Balanced equation including N_2 from air

The balanced equation ignoring N_2 from air is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2

Including N_2 from air, the balanced equation is

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2

(b) Balanced equation for 120 % stoichiometric combustion

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air

(d) Air:fuel mass ratio for 100 % combustion

Air:fuel = 17 300 kg/1700 kg = 10.2 :1

(e) Air:fuel mass ratio for 120 % combustion

Mass of air = 17 300 kg × 1.20 = 20 760 kg air

Air:fuel = 20 760 kg/1700 kg = 12.2 :1

6 0

3 years ago

What can we learn about past climate changes from ice cores?

Mice21 [21]

Fossil fuels about into Earth ice age time

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4 years ago

Why Lu* ion is highly hydrated?

LUCKY_DIMON [66]

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4 0

3 years ago

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Paha777 [63]

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3 0

3 years ago

A 10.0cm3 volume of alcohol has a mass of 7.05g. What is the density of the alcohol in grams per cubic centimeter

ExtremeBDS [4]

Answer:

0.705 g/cm³

Explanation:

Step 1: define density

The (volumetric mass) density of a sample or substance, is this sample or substance its mass per unit volume and can be defined through following formula:

ρ = m/v

with ρ = density (kg/m³)

m = mass of the substance (kg )

v = volume (m3)

Step 2: change datat SI-units

10.0 cm³ = 10^-5 m³

7.05 g = 0.00705 kg

ρ = 0.00705 kg / 10^-5 m³ = 705 kg / m³

To convert this to g/cm³ we have to divide by 1000

ρ = 0.705 g/cm³

To control this we calculate 7.05g / 10cm³ = 0.705 g/cm³

The desnity of the alcohol is 0.705 g/cm³

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4 years ago