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3 years ago

Please help balance _Ch7H16+_O2=_CO2+_H2O

Chemistry

1 answer:

iren2701 [21]3 years ago

8 0

Answer:

C7H16 + 11 O2 → 7 CO2 + 8 H2O

This is an oxidation-reduction (redox) reaction:

7 C-16/7 - 44 e- → 7 CIV (oxidation)

22 O0 + 44 e- → 22 O-II (reduction)

C7H16 is a reducing agent, O2 is an oxidizing agent.

Reactants:

C7H16

Names: Dioxygen source: wikidata, accessed: 2019-09-07, Oxygen source: ICSC, accessed: 2019-09-04source: wikidata, accessed: 2019-09-07, Oxygen (liquefied) source: ICSC, accessed: 2019-09-04

Appearance: Odourless compressed gas source: ICSC, accessed: 2019-09-04; Liquefied gas. colourless-to-blue extremely cold liquid source: ICSC, accessed: 2019-09-04

Products:

CO2

Names: Carbon dioxide source: wikipedia, accessed: 2019-09-27source: wikidata, accessed: 2019-09-02source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02, {{plainlist| source: wikipedia, accessed: 2019-09-27, CO2 source: wikidata, accessed: 2019-09-02

Appearance: Colorless gas source: wikipedia, accessed: 2019-09-27; Odourless colourless compressed liquefied gas source: ICSC, accessed: 2019-09-04; Colorless, odorless gas. [Note: Shipped as a liquefied compressed gas. Solid form is utilized as dry ice.] source: NIOSH NPG, accessed: 2019-09-02

H2O – Water, oxidane source: wikipedia, accessed: 2019-09-27

Other names: Water (H2O) source: wikipedia, accessed: 2019-09-27, Hydrogen hydroxide (HH or HOH) source: wikipedia, accessed: 2019-09-27, Hydrogen oxide source: wikipedia, accessed: 2019-09-27

Appearance: White crystalline solid, almost colorless liquid with a hint of blue, colorless gas source: wikipedia, accessed: 2019-09-27

Search by reactants (C7H16, O2)

1 O2 + C7H16 → H2O + CO2

2 O2 + C7H16 → H2O + CO

3 O2 + C7H16 → H2O + CO2 + CO

4 O2 + C7H16 → H2 + CO2

Search by products (CO2, H2O)

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1 HCl + CaCO3 → H2O + CO2 + CaCl2

2 HCl + NaHCO3 → H2O + CO2 + NaCl

3 O2 + CH4 → H2O + CO2

4 NaHCO3 → H2O + CO2 + Na2CO3

5 H2SO4 + KMnO4 + (COOH)2 → H2O + CO2 + K2SO4 + MnSO4

6 O2 + C3H8 → H2O + CO2

7 H2SO4 + K2CO3 → H2O + CO2 + K2SO4

8 O2 + C2H6 → H2O + CO2

9 CH3COOH + NaHCO3 → H2O + CO2 + CH3COONa

10 HCl + Na2CO3 → H2O + CO2 + NaCl

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Search by reactants (C7H16, O2) and by products (CO2, H2O)

1 O2 + C7H16 → H2O + CO2

2 O2 + C7H16 → H2O + CO2 + CO

Explanation:

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A 0.1873 g sample of a pure, solid acid, H2X (a diprotic acid) was dissolved in water and titrated with 0.1052 M NaOH solution.

puteri [66]

Answer:

We need 41.8 mL of NaOH

Explanation:

Step 1: Data given

Mass of H2X = 0.1873 grams

Molarity of NaOH solution = 0.1052 M

Molar mass of H2X = 85.00 g/mol

Step 2: The balanced equation

H2X (aq) +2 NaOH (aq) → Na2X (aq) + 2H2O(l)

Step 3: Calculate moles of H2X

Moles H2X = mass H2X / Molar mass H2X

Moles H2X = 0.1873 grams / 85.00 g/mol

Moles H2X = 0.0022 moles

Step 4: Calculate moles of NaOH

For 1 mol H2X we need 2 moles NaOH to produce 1 mole of Na2X and 2 moles of H2O

For 0.0022 moles of H2X we need 0.0044 moles of NaOH

Step 5: Calculate volume of NaOH

Volume of NaOH = moles of NaOH / molarity of NaOH

Volume of NaOH = 0.0044 moles / 0.1052 M

Volume NaOH = 0.0418 L = 41.8 mL

We need 41.8 mL of NaOH

6 0

3 years ago

The coolant in automobiles is often a 50/50 % by volume mixture of ethylene glycol, C2H6O2, and water. At 20°C, the density of e

Misha Larkins [42]

Explanation:

Let the volume of the solution be 100 ml.

As the volume of glycol = 50 = volume of water

Hence, the number of moles of glycol = $\frac{mass}{molar mass}$

= $\frac{density \times volume}{molar mass}$

= $\frac{1.1088 \times 50}{62 g/mol}$

= 0.894 mol

Hence, number of moles of water = $\frac{50 \times 0.998}{18}$

= 2.77

As glycol is dissolved in water.

So, the molality = $0.894 \times \frac{1000}{49.92}$

= 17.9

Therefore, the expected freezing point = $-1.86 \times 17.9$

= $-33.31^{o}C$

Thus, we can conclude that the expected freezing point is $-33.31^{o}C$ .

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3 years ago

Use standard enthalpies of formation to calculate ΔH∘rxn for the following reaction______.2H2S(g)+3O2(g)→2H2O(l)+2SO2(g) ΔH∘rxn

Ahat [919]

Answer:

-1,103.39KJ/mol

Explanation:

We use the subtract the standard enthalphies of formation of the reactants from that of the products. It must be taken into consideration that the enthalpy of formation of elements and their molecules alone are not taken into consideration. Hence, what we would be considering are the standard enthalpies of formation of H2S, H2O and SO2.

In places where we have more than one mole, we multiply by the number of moles as seen in the balanced chemical equations.

The standard enthalpies of the molecules above are as follows:

H2S = -20.63KJ/mol

H2O = -285.8KJ/mol

SO2 = -296.84KJ/mol

O2 = 0KJ/mol

ΔrH⦵ = [2ΔfH⦵(H2O) + 2 ΔfH⦵(SO2)] − [ΔfH⦵(H2S) + 3

 ΔfH⦵(O2)]

ΔrH⦵ =[(2 × -285.8) + (2 × -296.84)]

-[ 3 × -20.63)]

= (-571.6 - 593.68 + 61.89) = -1,103.39KJ/mol

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3 years ago

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Vilka [71]

Answer:

both forms of energy referred to in the question is light and heat energy

light energy is the visible energy that travels at a known constant speed of 3.0×10^9m/s

while heat energy is the invisible energy that travels in form of radiation at variable speeds

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2 years ago

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Kisachek [45]

Answer:

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Explanation:

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