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Mice21 [21]
3 years ago
7

Ethanol is added to gasoline because the oxygen it contains improves gasoline's burning efficiency. Its combustion reaction is g

iven below. CH3CH2OH(l) 3O2(g) → 2CO2(g) 3H2O(g) ∆H
A) Is this an exothermic or endothermic reaction? B) If 15.3 grams of pure ethanol is completely converted to products, calculate how much heat is absorbed or released, in (1) kJ and in (2) kcal. C) If 42.7 g of water vapor is produced, how much heat, in kl, is absorbed or released?
Chemistry
1 answer:
Angelina_Jolie [31]3 years ago
5 0

Answer:

A) Exothermic

B) (1) -410.5 kJ  (2) -98.1 kcal

C) -976.44 kJ

Explanation:

A) By the value of ΔH (-1236 kJ), we can see that it's negative, it means that the heat is being lost by the reaction, and, because of that, the reaction is exothermic.

B) The molar mass of ethanol is 46.07 g/mol, so, at 15.3 grams, the number of moles is:

n = mass/molar mass

n = 15.3/46.07

n = 0.3321 mol

By the equation

1 mol of ethanol           ----------------------- -1236 kJ

0.3321 mol of ethanol ----------------------- x

By a simple direct three rule:

x = -410.5 kJ

1 kJ --------------- 0.2390 kcal

-410 kJ ---------- y

y = -98.10 kcal

C) The molar mass of water is 18 g/mol, so the number of moles at 42.7 g is:

n = 42.7/18

n = 2.37 moles

By the equation

3 moles of water     ----------------------- -1236 kJ

2.37 moles of water --------------------- x

By a simple direct three rule:

3x = -2929.32

x = -976.44 kJ

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Answer:

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5 0
3 years ago
When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were
uysha [10]

Answer:

\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}

(b) Mass of H

\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}

(c)Mass of Fe

(i)In 0.4131g of X

\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}

(ii) In 1.2383 g of X

\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}

5 0
3 years ago
Why are familiar objects such as pens and paper clips not commonly counted in moles?
Anton [14]
Well ask yourself why don't we count it in moles and you should get your answer.
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3. Which of the following statements best describes friction?
mamaluj [8]

Answer:

d.

Explanation:

It is a force that occurs when surfaces touch each other.

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2 years ago
According to the vsepr model the arrangement of electron pairs around nh3 and ch4 is
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3 years ago
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