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2 years ago

Benzene, C6H6, reacts with oxygen, O2, to form CO2 and H2O. How much O2 is required for the complete combustion of 1.0 mol C6H6?

2 answers:

4 0

C6H6 + 02 forms CO2 + H20
Complete with factors to stabilize
C6H6 + 15/2 O2 forms 6 CO2 + 3 H2O
You take away only the info you need
C6H6 15/2 O2
1 mol 15/2 or 7,5 mol

15/2 or 7,5 mol of O2 are required.

;)

Monica [59]2 years ago

3 0

Answer:

7.5 mol of oxygen are required

Explanation:

Hi, as it says, this is a combustion reaction of benzene:

$C_6H_6(l) + O_2(g) \longrightarrow CO_2 (g) + H_2O (g)$

The first step is to balance the equation:

1 mol of $CO_2$ for each C in the benzene

1 mol of $H_2O$ for each two H in the benzene

1 mol of $O_2$ for each $CO_2$ and 0.5 mol for each $H_2O$

Balanced:

$C_6H_6(l) + 15/2 O_2(g) \longrightarrow 6 CO_2 (g) + 3 H_2O (g)$

Now, following the reaction:

For 1 mol of benzene, 15/2 (7.5) mol of oxygen are required.

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Calculate the amount, in moles, of PO43- present at equilibrium when excess Sr3(PO4)2 is added to 750. mL 1.2 M Sr(NO3)2(aq). As

Crank

Answer:

1.8 × 10⁻¹⁶ mol

Explanation:

(a) Calculate the solubility of the Sr₃(PO₄)₂

Let s = the solubility of Sr₃(PO₄)₂.

The equation for the equilibrium is

Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹

1.2 + 3s 2s

$K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} = 1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\$

(b) Concentration of PO₄³⁻

[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹

(c) Moles of PO₄³⁻

Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol

7 0

2 years ago

I need help Type th temperature (in Kelvin) in the left column and the volumes in the right column, being sure to keep pairs of

e-lub [12.9K]

Answer:

sorry nahi pata tha muje

7 0

2 years ago

Read 2 more answers

Ethylene (C2H4) is the starting material for the preparation of polyethylene. Although typically made during the processing of p

Solnce55 [7]

Answer:

1.17 grams

Explanation:

Let's consider the balanced equation for the combustion of ethylene.

C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(l)

We can establish the following relations:

1411 kJ are released (-1411 kJ) when 1 mole of C₂H₄ burns.
The molar mass of C₂H₄ is 28.05 g/mol.

The grams of C₂H₄ burned to give 59.0 kJ of heat (q = -59.0 kJ) is:

$-59.0kJ.\frac{1molC_{2}H_{4}}{-1411kJ} .\frac{28.05gC_{2}H_{4}}{1molC_{2}H_{4}} =1.17gC_{2}H_{4}$

8 0

2 years ago

Pesticide concentrations in the Rhine River between Germany and France between 1969 and 1975 averaged 0.55 mg/L of hexachloroben

sertanlavr [38]

Answer:

1.93×10⁻³ mmoles/L of C₆Cl₆; 1.58×10⁻⁴ mmoles/L of C₁₂H₈Cl₆O; 3.51×10⁻³ mmoles/L of C₆H₆Cl₆

Explanation:

We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

C₆Cl₆ → 12. 6 + 35.45 .6 = 284.7 g/m

C₁₂H₈Cl₆O → 12 . 12 + 8.1 + 35.45 .6 + 16 = 380.7 g/m

C₆H₆Cl₆ → 12.6 + 6.1 + 35.45 .6 = 290,7 g/m

Let's convert mg to g (/1000)

0.55 mg / 1000 = 5.5×10⁻⁴ g

0.060 mg / 1000 = 6×10⁻⁵ g

1.02 mg / 1000 = 1.02×10⁻³ g

Now we can know the moles (mass / molar mass)

5.5×10⁻⁴ g / 284.7 g/m = 1.93×10⁻⁶ moles of C₆Cl₆

6×10⁻⁵ g / 380.7 g/m = 1.58×10⁻⁷ moles of C₁₂H₈Cl₆O

1.02×10⁻³ g / 290,7 g/m = 3.51×10⁻⁶ moles of C₆H₆Cl₆

Milimoles = Mol . 1000

1.93×10⁻⁶ . 1000 = 1.93×10⁻³ mmoles of C₆Cl₆

1.58×10⁻⁷ . 1000 = 1.58×10⁻⁴ mmoles of C₁₂H₈Cl₆O

3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

6 0

2 years ago

Convert 2.0 M of Phenobarbital sodium (MW: 254 g/mole) solution in water into % w/v and ratio strengths.

Travka [436]

Answer:

The concentration is 50,8 % w/v and radio strengths = 1,96.

Explanation:

Phenobarbital sodium is a medication that could treat insomnia, for example.

2,0 M of Phenobarbital sodium means 2 moles in 1L.

The concentration units in this case are %w/v that means 1g in 100 mL and ratio strengths that means 1g in <em>r</em> mL. Thus, 2 moles must be converted in grams with molar weight -254 g/mole- and liters to mililiters -1 L are 1000mL-. So:

2 moles × $\frac{254 g}{1 mole}$ = 508 g of Phenobarbital sodium.

1 L × $\frac{1000 mL}{ 1 L}$ = 1000 mL of solution

Thus, % w/v is:

$\frac{508 g}{1000 mL}$ × 100 = 50,8 % w/v

And radio strengths:

$\frac{1000 mL}{508 g}$ = 1,96. Thus, you have 1 g in 1,96 mL

I hope it helps!

5 0

2 years ago