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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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What mass of Fe(OH)3 is produced when 35 mL of 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M
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Answer:

0.35 g.

Explanation:

We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.

This is illustrated below:

Molarity of Fe(NO3)3 = 0.250 M

Volume = 35 mL = 35/1000 = 0.035 L

Mole of Fe(NO3)3 =?

Molarity = mole /Volume

0.250 = mole of Fe(NO3)3 / 0.035

Cross multiply

Mole of Fe(NO3)3 = 0.25 x 0.035

Mole of Fe(NO3)3 = 8.75×10¯³ mole

Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M

KOH solution. This is illustrated below:

Molarity of KOH = 0.180 M

Volume = 55 mL = 55/1000 = 0.055 L

Mole of KOH =.?

Molarity = mole /Volume

0.180 = mole of KOH /0.055

Cross multiply

Mole of KOH = 0.180 x 0.055

Mole of KOH = 9.9×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

3KOH + Fe(NO3)3 —> Fe(OH)3 + 3KNO3

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3.

Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.

From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.

Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.

Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.

The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.

Finally, we shall convert 3.3×10¯³ mole of Fe(OH)3 to grams. This can be obtained as follow:

Molar mass of Fe(OH)3 = 56 + 3(16 + 1) = 56 + 3(17) = 107 g/mol

Mole of Fe(OH)3 = 3.3×10¯³ mole

Mass of Fe(OH)3 =?

Mole = mass /Molar mass

3.3×10¯³ = Mass of Fe(OH)3 / 107

Cross multiply

Mass of Fe(OH)3 = 3.3×10¯³ x 107

Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.

Therefore, 0.35 g of Fe(OH)3 was produced from the reaction.

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