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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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hope this helps !.



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4 0
3 years ago
You have a solution of 600 mg of caffeine dissolved in 100 mL of water. The partition coefficient for aqueous caffeine extracted
klio [65]

Answer:

159 mg caffeine is being extracted in 60 mL dichloromethane

Explanation:

Given that:

mass of caffeine in 100 mL of water =  600 mg

Volume of the water = 100 mL

Partition co-efficient (K) = 4.6

mass of caffeine extracted = ??? (unknown)

The portion of the DCM = 60 mL

Partial co-efficient (K) = \frac{C_1}{C_2}

where; C_1= solubility of compound in the organic solvent and C_2 = solubility in aqueous water.

So; we can represent our data as:

K=(\frac{A_{(g)}}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL} )

Since one part of the portion is A and the other part is B

A+B = 60 mL

A+B = 0.60

A= 0.60 - B

4.6=(\frac{0.6-B(mg)}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL})

4.6 = \frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}

4.6 × (\frac{B_{(mg)}}{100mL}) = (\frac{0.6-B(mg)}{60mL} )

4.6 B *\frac{60}{100} = 0.6 - B

2.76 B = 0.6 - B

2.76 + B = 0.6

3.76 B = 0.6

B = \frac{0.6}{3.76}

B = 0.159 g

B = 159 mg

∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.

4 0
3 years ago
Read 2 more answers
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