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Harman [31]
3 years ago
11

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

O2

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)

12345678910...

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

12345678910...

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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From the question,

Applying

PV/T = P'V'/T'......................... Equation 1

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V' = PVT'/P'T......................... Equation 2

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3 years ago
What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2
atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}

Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

Mole of AlCl_3 = Mole of Al * Mole ratio

Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = 133.34 \frac{g}{mol}

Mass of AlCl3 can be calculated using mole formula as:

1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}

Multiplying both side by 133.34 \frac{g}{mol}

1.32 mole  * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34  \frac{g}{mol}

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

Mole of AlCl_3 = mole of Cl_2 * mole ratio

Mole of AlCl_3 = 0.55  mole  * \frac{2}{3}

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}

Multiplying both side by

133.34 \frac{g}{mol}

0.367 mol of AlCl_3 * 133.34 \frac{g}{mol}  = \frac{mass(g)}{133.34\frac{g}{mol}}   * 133.34 \frac{g}{mol}

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

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There's only one electron in hydrogen.

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