What happens when the temperature of a solid reaches it’s melting point

If 28 grams of N reacts completely with 12 grams of H2, then how many

Bogdan [553]

Answer:

Mass of NH₃ produced = 34 g

Explanation:

Given data:

Mass of nitrogen = 28 g

Mass of Hydrogen = 12 g

Mass of NH₃ produced = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 28 g/ 28 g/mol

Number of moles = 1 mol

Moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 12 g/ 2 g/mol

Number of moles = 6 mol

Now we will compare the moles of hydrogen and nitrogen with ammonia.

H₂ : NH₃

3 : 2

6 : 2/3×6 = 4 mol

N₂ : NH₃

1 : 2

Number of moles of ammonia produced by nitrogen are less thus it will act as limiting reactant.

Mass of ammonia produced:

Mass = number of moles × molar mass

Mass = 2 mol × 17 g/mol

Mass = 34 g

5 0

3 years ago

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Which of the following is an example of an exothermic reaction? silt settling out of muddy water, bubbles forming in soda, photo

Rudik [331]

Fireworks
An exothermic reaction is one where the products have lower energy than the reactants, so the reaction yields energy. The chemical compounds present in firework fuel release a lot of energy upon oxidation. Photosynthesis is endothermic, settling of silt is not a chemical reaction, it is a physical change. Finally, the bubble formation in soda is not exothermic; otherwise, the sodas would become very hot very fast.

4 0

3 years ago

What is the oxidation number of HCLO4?

ki77a [65]

E.g. in H3PO4 (O, -2).
8. The sum of the oxidation states of all the atoms in a species must be equal to net charge on the species. e.g. Net Charge of HClO4 = 0, i.e. [+1(H)+7(Cl)-2*4(O)] = 0.

4 0

3 years ago

List three quantum numbers you know

Serhud [2]

Answer:principle quantum number

Angular quantum number

Magnetic quantum number

Spin quantum number

Explanation:

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3 years ago

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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;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(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

(c) Minimum mass of air

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