Answered: In 1997, at the United Nations…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

In 1997, at the United Nations Conference on Climate Change, the major industrial nations agreed to expand their re-search efforts to develop renewable sources of carbon-basedfuels. For more than a decade, Brazil has been engaged in aprogram to replace gasoline with ethanol derived from the rootcrop manioc (cassava).(a) Write separate balanced equations for the complete combus-tion of ethanol (C₂H₅OH) and of gasoline (represented by theformula C₈H₁₈).(b) What mass of oxygen is required to burn completely 1.00 Lof a mixture that is 90.0% gasoline (d=0.742 g/mL) and 10.0%ethanol (d=0.789 g/mL) by volume?(c) If 1.00 mol of O₂ occupies 22.4 L, what volume of O₂ isneeded to burn 1.00 L of the mixture?(d) Air is 20.9% O₂ by volume. What volume of air is needed toburn 1.00 L of the mixture?

Expert Solution

Step 1

As per our guidelines we have allowed to answer the first three subpart for you. Kindly repost the rest as separate question.

(a)

The separate equation for the complete combustion reaction of ethanol is given as,

$C_{2} H_{5} O H + 3 O_{2} \overset{}{\to} 2 C O_{2} + 3 H_{2} O$

The separate equation for the complete combustion reaction of gasoline is given as,

$2 C_{8} H_{18} + 25 O_{2} \overset{}{\to} 16 C O_{2} + 18 H_{2} O$

Step 2

(b) The mass of oxygen required is calculated as,

The complete combustion reaction of ethanol is given as,

$C_{2} H_{5} O H + 3 O_{2} \overset{}{\to} 2 C O_{2} + 3 H_{2} O$

The density of ethanol is $0.789 g / m L$ .

The volume of ethanol is $0.1 L$ .

The mass of ethanol is calculated as,

$M a s s = V o l u m e \times D e n s i t y M a s s = 0.1 L \times 0.789 g / m L m = 78.9 g$

The mass of ethanol is $78.9 g$ .

The moles of ethanol is calculated using the mass and molar mass of ethanol.

$M o l e s = \frac{78.9 g}{46.1 g / m o l} M o l e s = 1.713 m o l$

The moles of ethanol is $1.713 m o l$ .

From the moles of ethanol, moles of oxygen is calculated as,

Moles of oxygen= $3 \times (M o l e s o f C_{2} H_{5} O H)$

Moles of oxygen= $3 \times (1.713 m o l)$

Moles of oxygen=5.139mol

The moles of oxygen from moles of ethanol is 5.139mol.

Step 3

The separate equation for the complete combustion reaction of gasoline is given as,

$2 C_{8} H_{18} + 25 O_{2} \overset{}{\to} 16 C O_{2} + 18 H_{2} O$

The density of gasoline is $0.742 g / m L$ .

The volume of gasoline is $0.9 L$ .

The mass of gasoline is calculated as,

$M a s s = V o l u m e \times D e n s i t y M a s s = 0.9 L \times 0.742 g / m L m = 667.8 g$

The mass of gasoline is $667.8 g$ .

The moles of gasoline is calculated using the mass and molar mass of gasoline

$M o l e s = \frac{667.8 g}{114.23 g / m o l} M o l e s = 5.846 m o l$

The moles of gasoline is $5.846 m o l$ .

From the moles of gasoline, moles of oxygen is calculated as,

Moles of oxygen= $25 / 2 (m o l e s o f C_{8} H_{18})$

Moles of oxygen= $25 / 2 (5.846 m o l)$

Moles of oxygen=73.076mol

The moles of oxygen from moles of gasoline is 73.076mol.

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