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Alex_Xolod [135]
3 years ago
11

For the combustion reaction of C9H12 in O2: how many moles of O2 is required to react with 0.67 mol C9H12?

Chemistry
2 answers:
miskamm [114]3 years ago
7 0

Answer:

8.0 mol O₂

Explanation:

Let's consider the complete combustion reaction of C₉H₁₂.

C₉H₁₂ + 12 O₂ → 9 CO₂ + 6 H₂O

The molar ratio of C₉H₁₂ to O₂ is 1:12. The moles of O₂ required to react with 0.67 moles of C₉H₁₂ are:

0.67 mol C₉H₁₂ × (12 mol O₂/1 mol C₉H₁₂) = 8.0 mol O₂

8.0 moles of O₂ are required to completely react with 0.67 moles of C₉H₁₂.

GenaCL600 [577]3 years ago
6 0

Answer:

To react with 0.67 moles C9H12 we need 8.04 moles of O2

Explanation:

Step 1: Data given

Number of moles C9H12 = 0.67 moles

Step 2: The balanced equation

C9H12 + 12O2 → 9CO2 + 6H2O

Step 3: Calculate moles of O2 required

For 1 mol C9H12 we need 12 moles of O2 to produce 9 moles of CO2 and 6 moles of H2O

For 0.67 moles of C9H12 we need 12 *0.67 = 8.04 moles of O2

To produce 9*0.67 = 6.03 moles of CO2 and 6*0.67 = 4.02 moles H2O

To react with 0.67 moles C9H12 we need 8.04 moles of O2

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The concentration of iron in ppb is 5.2 ppb

In atmospheric chemistry, PPM( parts per million) and PPB( parts per billion) units are used to express the concentration of gases. PPB stands for parts per billion, while PPM stands for parts of gas per million parts of air.

2.6 × 10⁻⁶ grams of iron in 500 ml of water.

Now, converting milliliter into liter:

500 mL = 500 mL × ( 1L / 1000 mL ) = 0.5 L

Now for ppm, mass must be in mg.

2.6 × 10⁻⁶ grams = 2.6 × 10⁻⁶ grams × ( 1000 mg / 1 grams ) = 2.6 × 10⁻³ mg

Then, the concentration of iron in ppm will be:

2.6 × 10⁻³ mg / 0.5 L = 5.2 × 10⁻³ ppm

Now, 1 ppm = 1000 ppb

Therefore,

5.2 × 10⁻³ ppm × 10³ = 5.2 ppb

Learn more about concentration here:

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8 0
9 months ago
Why is potassium nitrate classified as an electrolyte?.
ch4aika [34]

Answer: It is a molecular compound. (2) It contains a metal. (3) It can conduct electricity as a solid.

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2 years ago
A solution is prepared from 4.5701 g of magnesium chloride and 43.238 g of water. The vapor pressure of water above this solutio
balandron [24]

Answer:

i = 2.483

Explanation:

The vapour pressure lowering formula is:

Pₐ = Xₐ×P⁰ₐ <em>(1)</em>

For electrolytes:

Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ

Where:

Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)

4.5701g of MgCl₂ are:

4.5701g ₓ (1mol / 95.211g) = 0.048000 moles

43.238g of water are:

43.238g ₓ (1mol / 18.015g) = 2.400 moles

Replacing in (1):

0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm

0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)

2.4mol + i*0.048mol = 2.4mol / 0.9527

2.4mol + i*0.048mol = 2.5192mol

i*0.048mol = 2.5192mol - 2.4mol

i = 0.1192mol / 0.048mol

<em>i = 2.483</em>

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I hope it helps!

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