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Marta_Voda [28]
3 years ago
12

Write a balanced equation for the complete combustion of 2,3-dimethylbutane. use the molecular formula for the alkane (c before

h) and the smallest possible integer coefficients.
Chemistry
2 answers:
Pachacha [2.7K]3 years ago
6 0

Answer:

The balanced chemical reaction is given as:

2C_6H_{14}+19O_2\rightarrow 12CO_2+14H_2O

Explanation:

Combustion is a type of chemical reaction in which hydrocarbon reacts with oxygen gas to form carbon dioxide gas and water .

When 2,3-dimethylbutane undergoes combustion reaction it gives carbon dioxde and water.

The balanced chemical reaction is given as:

2C_6H_{14}+19O_2\rightarrow 12CO_2+14H_2O

According to stoichiometry, 2 moles of 2,3-dimethylbutane reacts with 19 moles of oxygen gas to give 12 moles of carbon dioxide and 14 moles of water.

Nuetrik [128]3 years ago
4 0

2C_6H_14 + 19O_2 → 12CO_2 + 14H_2O

<em>Step 1</em>. Write the <em>condensed structural  formula</em> for 2,3-dimethylbutane.

(CH_3)_2CHCH(CH_3)_2

<em>Step 2</em>. Write the <em>molecular formula</em>.

C_6H_14

<em>Step 3</em>. Write the <em>unbalanced chemical equation</em>.

C_6H_14 + O_2 → CO_2 + H_2O

<em>Step 4</em>. Pick the <em>most complicated-looking formula</em> (C_6H_14) and balance its atoms (C and H).

<em>1</em>C_6H_14 + O_2 → <em>6</em>CO_2 + <em>7</em>H_2O

<em>Step 5</em>. Balance the <em>remaining atoms</em> (O).

1C_6H_14 + (<em>19/2</em>)O_2 → 6CO_2 + 7H_2O

Oops! <em>Fractional coefficients</em>!

<em>Step 6</em>. <em>Multiply all coefficients by a number</em> (2) to give integer coeficients..

2C_6H_14 + 19O_2 → 12CO_2 + 14H_2O

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4 years ago
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How many molecules of F2 react with 66.6 g NH3 ? (stoichiometry)
stiks02 [169]
<h3>Answer:</h3>

5.89 × 10^23 molecules of F₂

<h3>Explanation:</h3>

The equation for the reaction between fluorine (F₂) and ammonia (NH₃) is given by;

5F₂ + 2NH₃ → N₂F₄ + 6 HF

We are given 66.6 g NH₃

We are required to determine the number of fluorine molecules

<h3>Step 1: Moles of Ammonia </h3>

Moles = Mass ÷ Molar mass

Molar mass of ammonia = 17.031 g/mol

Moles of NH₃ = 66.6 g ÷ 17.031 g/mol

                      = 3.911 moles

<h3>Step 2: Moles of Fluorine </h3>

From the equation 5 moles of Fluorine reacts with 2 moles of ammonia

Therefore,

Moles of fluorine = Moles of Ammonia × 5/2

                            = 3.911 moles × 5/2

                           = 9.778 moles

<h3>Step 3: Number of molecules of fluorine </h3>

We know that 1 mole of a compound contains number of molecules equivalent to the Avogadro's number, 6.022 × 10^23 molecules

Therefore;

1 mole of F₂ = 6.022 × 10^23 molecules

Thus,

9.778 moles of F₂ = 9.778 moles × 6.022 × 10^23 molecules/mole

                              = 5.89 × 10^23 molecules

Therefore, the number of fluorine molecules needed is 5.89 × 10^23 molecules

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