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Mnenie [13.5K]
2 years ago
10

Given the reaction below, how many grams of Li3N can be formed from 6.07 g of Li? Assume an excess of nitrogen.

Chemistry
1 answer:
Sonbull [250]2 years ago
8 0

Answer:

About 10.2 g Li₃N.

Explanation:

We are given the reaction:

\displaystyle \text{6 Li(s) + N$_2$(g) }\longrightarrow \text{2 Li$_3$N(s)}

And we want to determine the amount of Li₃N that can be formed from 6.07 g of Li and an excess of nitrogen.

To convert from g Li to g Li₃N, we can: (1) convert from g Li to mol Li, (2) mol Li to mol Li₃N, and (3) mol Li₃N to g Li₃N.

  1. The molecular weight of Li is 6.94 g/mol.
  2. From the equation, six moles of Li yields two moles of Li₃N.
  3. And the molecular weight of Li₃N is 34.83 g/mol as shown below.

Molecular weight of Li₃N:

\displaystyle \begin{aligned}\text{MW}_\text{Li$_3$N} & = (3 (6.94) + 14.01) \text{ g/mol} \\ \\ & =34.83\text{ g/mol} \end{aligned}

This yields three ratios:

\displaystyle \frac{1 \text{ mol Li}}{6.94 \text{ g Li}}, \frac{2 \text{ mol Li$_3$N}}{6 \text{ mol Li}}, \text{ and } \frac{34.83 \text{ g Li$_3$N}}{1 \text{ mol Li$_3$N}}

From the initial value, multiply:

\displaystyle 6.07 \text{ g Li} \cdot \frac{ 1 \text{ mol Li}}{6.94 \text{ g Li}} \cdot \frac{2 \text{ mol Li$_3$N}}{6 \text{ mol Li}} \cdot \frac{34.83 \text{ g Li$_3$N}}{1 \text{ mol Li$_3$N}} = 10.2\text{ g Li$_3$N}

In conclusion, 10.2 g of Li₃N is formed from 6.07 g of Li and an excess of nitrogen.

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Answer:

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<u>Ptolemy's model:</u>

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Let's begin by the fact both models are geocentric  and make a brief explanation of them:

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So, according to this model, the universe was spherical and finite, with the Earth immobile at its center, composed of the four fundamental elements (made up of spherical layers): <em>ground, water, air and fire</em>; and the Sun along with the fixed planets in their respective concentric spheres (also called nested spheres) revolving around the Earth.

<u>Ptolemy's model:</u>

Ptolemy's model of the universe was also geocentric, placing the Earth motionless in the center of the known universe and was the accepted model for a long time during the Middle Ages.

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However, this model predicted with an acceptable accuracy for that time the motion of the known planets, until Copernicus appeared with the revolutionary Heliocentric model.

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Note down the formula below

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Mass % of Hydrogen:-

\\ \sf\longmapsto \dfrac{15}{135}\times 100

\\ \sf\longmapsto \dfrac{1}{9}\times 100

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Mass % of Oxygen:-

\\ \sf\longmapsto \dfrac{120}{135}\times 100

\\ \sf\longmapsto \dfrac{8}{9}\times 100

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<u>Explanation:</u>

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