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2 years ago

How many liters of oxygen are required to react completely with 14.8 mol of Al?

Chemistry

1 answer:

Bond [772]2 years ago

8 0

Answer:

296 L

Explanation:

We will need a balanced equation with moles, so let's gather all the information in one place.

4Al + 3O₂ ⟶ 2Al₂O₃

n/mol: 17.4

1. Moles of O₂

$n = \text{17.4 mol Al}\times \dfrac{\text{3 mol O}_{2}}{\text{4 mol Al}}= \text{13.05 mol O}_{2}$

2. Volume of O₂

You haven't given the conditions at which the volume is measured, so I assume it is at STP (0 °C and 1 bar).

At STP the molar volume of a gas is 22.71 L.

$V = \text{13.05 mol}\times \dfrac{\text{22.71 L}}{\text{1 mol }}= \textbf{296 L}$

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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

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Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is:

$s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}$

$s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}$

The change in entropy is -1083.112 joules per kilogram-Kelvin.

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