What is heat? O Fire O Hot O Energy O Potential

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

What is the molarity of a solution containing 501mL with 35g NaCl

Vesna [10]

Answer:

Data:

mass of solute: 35g of NaCl

m.mass of solute: 58g/mol

volume of solution: 501mL

Molarity=?

Explanation:

501ml = 0.5dm3

M= g of solute/m.mass of solute*vol of solution

M= 35/58*0.5

M=1.20

6 0

3 years ago

What mass of solid that has a molar mass of 46.0 g/mol should be added to 150.0 g of benzene to raise the boiling point of benze

enyata [817]

Answer : 17.12 g

Explanation: $\Delta T =k_b\times m$

$\Delta T$ = elevation in boiling point

$k_b$ = boiling point elevation constant

m= molality

$molality=\frac{mass of solute}{molecular mass of solute\times weight of the solvent in kg}$

given $\Delta T=6.28^{\circ}C$

Molar mass of solute = 46.0 $gmol^{-1}$

Weight of the solvent = 150.0 g = 0.15 kg

Putting in the values

$molality=\frac{x}{46gmol^{-1}\times0.15kg}$

$6.28 =2.53^{\circ}Ckgmol^{-1}\frac{x}{46gmol^{-1}\times 0.15kg}$

x = 17.12 g

3 0

3 years ago

If I need 2.2 moles of CO2 , and I have excess Fe2O3 , how many moles of C do I need?

olchik [2.2K]

Answer:

0.733 mol.

Explanation:

From the balanced equation:

2Fe₂O₃ + C → Fe + 3CO₂,

It is clear that 1.0 moles of Fe₂O₃ react with 1.0 mole of C to produce 1.0 mole of Fe and 3.0 moles of CO₂.

Since Fe₂O₃ is in excess, C will be the limiting reactant.

Using cross multiplication:

1.0 mole of C produces → 3.0 moles of CO₂, from the stichiometry.

??? mole of C produces → 2.2 moles of CO₂.

∴ The no. of moles of C needed to produce 2.2 moles of CO₂ = (1.0 mole of C) (2.2 mole of CO₂) / (3.0 mole of CO₂) = 0.733 mol.

6 0

2 years ago

What mass of boron sulfide must be processed with 2.1 x 10 4g of carbon to yield 3.11 x 10 4 g of boron and 1.47 x 10 5 g of car

Basile [38]

The reaction between boron sulfide and carbon is given as:

2B2S3 + 3C → 4B + 3CS2

As per the law of conservation of mass, for any chemical reaction the total mass of reactants must be equal to the total mass of the products.

Given data:

Mass of C = 2.1 * 10^ 4 g

Mass of B = 3.11*10^4 g

Mass of CS2 = 1.47*10^5

Mass of B2S3 = ?

Now based on the law of conservation of mass:

Mass of B2S3 + mass C = mass of B + mass of CS2

Mass of B2S3 + 2.1 * 10^ 4 = 3.11*10^4 + 1.47*10^5

Mass of B2S3 = 15.7 * 10^4 g

4 0

3 years ago