Which proffesion is likely to use a radiation shield?

Please help I need the answer now. Write a balanced equation for combustion of pentane in plenty supply of air and in limited su

Ratling [72]

Pentane burns in plenty of air: CO₂ and H₂O is produced.

C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

Pentane burns in limited amount of air: CO or even C is produced along with H₂O.

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

<h3>Explanation</h3>

Pentane is a hydrocarbon. There are five carbon atoms in each of its molecule. Its molecular formula will be C₅H₁₂.

Hydrocarbon fuels burn to produce CO₂ when there's plenty of air.

? C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

Among all species in this reaction, C₅H₁₂ has the largest number of atoms per molecule. Assume that the coefficient of C₅H₁₂ is one.

1 C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

C₅H₁₂ is the only reactant that contains C atoms. There are 5 C atoms in a C₅H₁₂ molecule. There should be the same number of C atoms in the products.
CO₂ is the only product that contains C atoms. There are one C atom in each CO₂ molecule. 5 C atoms correspond to 5 CO₂ molecules.

1 C₅H₁₂ + ? O₂ → 5 CO₂ + ? H₂O

Similarly, C₅H₁₂ is the only reactant that contains H atoms. H₂O is the only product that contains H atoms. There are 12 H atoms in one C₅H₁₂ molecule, which corresponds to 6 H₂O molecules.

1 C₅H₁₂ + ? O₂ → 5 CO₂ + 6 H₂O

Both CO₂ and H₂O are products that contains O atoms. There are 5 × 2 + 6 × 1 = 16 O atoms in total in 5 CO₂ molecules and 6 H₂O molecules. The 16 O atoms on the product side corresponds to 8 O₂ molecules on the reactant side.

1 C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

1 C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

All coefficients shall be whole numbers. If there's any fraction in this equation, multiply both sides by the least common multiple of their denominators.

Hydrocarbon fuels burn to produce H₂O and CO when there's a limited supply of air. C (soot) might also be produced. Assuming that only CO is produced. Try to balance the equation using the same method.

1 C₅H₁₂ + 11/2 O₂ → 5 CO + 6 H₂O

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

Less O₂ is consumed for each mole of C₅H₁₂.

Consider: What would be the balanced equation when only C is produced?

<h3>Reference</h3>

"Products and effects of combustion", GCSE Chemistry (Single Science), BBC Bitesize.

5 0

3 years ago

Acid-catalyzed dehydration of an alcohol is an equilibrium situation. How was the reaction forced to completion in Acid-Catalyze

egoroff_w [7]

Answer:

In the acid-catalyzed dehydration of 2-methyl-2-butanol, the reaction can be driven to completion using Le Chatelier's principle. The reaction is driven to completion because the released water molecules form a strong bond with the acid used as a catalyst. As a result, the alkene produced can be distilled from the mixture.

Explanation:

In the acid-catalyzed dehydration of 2-methyl-2-butanol, the reaction can be driven to completion using Le Chatelier's principle. The reaction is driven to completion because the released water molecules form a strong bond with the acid used as a catalyst. As a result, the alkene produced can be distilled from the mixture.

3 0

3 years ago

If a decrease in temperature accompanies a reaction, what occurred?

marusya05 [52]

Released my friending

8 0

3 years ago

Read 2 more answers

How many grams of hydrogen are contained in 2.00 mol of C6H7N

Setler [38]

Answer:- 14.0 moles of hydrogen present in 2.00 moles of $[tex]C_6H_7N$ .

Solution:- We have been given with 2.00 moles of $C_6H_7N$ and asked to calculate the grams of hydrogen present in it. It's a two step conversion problem. In first step we convert the moles of the compound to moles of hydrogen as one mol of the compound contains 7 moles of hydrogen. In next step the moles are converted to grams on multiplying the moles by atomic mass of H. The calculations are shown as:

$2.00molC_6H_7N(\frac{7molH}{1molC_6H_7N})(\frac{1.0gH}{1molH})$

= 14.0 g H

So, there are 14.0 g of hydrogen in 2.00 moles of $C_6H_7N$ .

3 0

3 years ago

If 11g of a gas occupies 5.6dm'3at s.t.p., calculate it's vapour density (1.0mol of a gas occupies 22.4dm'3at s.t.p.).

11111nata11111 [884]

Answer:

${ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 = \frac{11}{m _{r}} \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\ \\ { \bf{vapour \: density = 2 \times m _{r}}} \\ = 2 \times 14.85 \\ = 29.7 \: { \tt{g {dm}^{ - 3} }}$

7 0

3 years ago