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Artemon [7]
3 years ago
6

g Liquid hexane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . What is the theoretical yie

ld of water formed from the reaction of 6.89 g of hexane and 47.4 g of oxygen gas
Chemistry
1 answer:
KIM [24]3 years ago
7 0

Answer:

10.09g

Explanation:

Step 1:

The balanced equation for the reaction.

We'll begin by writing a balanced equation for the reaction. This is illustrated below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Step 2:

Determination of the masses of C6H14 and O2 that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

Molar Mass of C6H14 = (12x6) + (14x1) = 72 + 14 = 86g/mol

Mass of C6H14 from the balanced equation = 2 x 86 = 172g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 19 x 32 = 608g

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 14 x 18 = 252g

Summary:

From the balanced equation above,

172g of C6H14 reacted with 608g of O2 to produce 252g of H2O.

Step3:

Determination of the limiting reactant.

It essential that we determine the limiting reactant because it will be use to determine the theoretical yield.

From the balanced equation above,

172g of C6H14 reacted with 608g of O2.

Therefore, 6.89g of C6H14 will react with = (6.89 x 608)/172 = 24.36g of O2.

We can see that there are left over for O2 as only 24.36g of O2 reacted out of 47.4g that was given. Therefore, C6H14 is the limiting reactant.

Step 4:

Determination of the theoretical yield of water from the reaction.

The limiting reactant will be used to obtain theoretical yield of water. This is illustrated below:

From the balanced equation above,

172g of C6H14 produce 252g of H2O.

Therefore, 6.89g of C6H14 will produce = (6.89 x 252)/172 = 10.09g of H2O.

Therefore, the theoretical yield of water (H2O) is 10.09g

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2 years ago
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Answer:

0.56L

Explanation:

This question requires the Ideal Gas Law:  PV=nRT where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.

Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:R=0.0821\frac{L\cdot atm}{mol\cdot K}

Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means T=273.15K and P=1atm

Lastly, we must calculate the number of moles of O_2(g) there are.  Given 0.80g of O_2(g), we will need to convert with the molar mass of O_2(g).  Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2:  32g\text{ }O_2=1mol\text{ }O_2

Thus, \frac{0.80g \text{ }O_2}{1} \frac{1mol\text{ }O_2}{32g\text{ }O_2}=0.25mol\text{ }O_2=n

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PV=nRT

V=\frac{nRT}{P}

...substituting the known values, and simplifying...

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Calculation,

Given information

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Formula used:

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By putting the valur of R,T, Kf we get the value of ΔG°

ΔG° = - 8.314 J/K• mol×300K㏑ 1.7 × 10^{7}

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Answer:

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Explanation:

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