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murzikaleks [220]
4 years ago
15

A 60g sample of tetraethyl lead, a gasoline additive, is found to contain 38.43g lead, 17.83g carbon and 3.74g hydrogen. Determi

ne the compounds empirical formula.
Chemistry
1 answer:
anastassius [24]4 years ago
6 0

Answer:

Empirical formula (which matches the molecular formula) is = PbC₈H₂₀

Explanation:

Our sample: 60 g of tetraethyl lead

In order to determine the compound empirical formula we need the centesimal composition:

(Mass of element / Total mass) . 100 =

(38.43 g lead / 60g ) . 100 = 64.05%

(17.83 g C / 60g) . 100 = 29.72%

(3.74 g H / 60g) . 100 = 6.23 %

These % are the mass of the elements in 100 g of compound. Let's find out the moles of them:

64.05 g / 207.2 g/mol = 0.309 moles

29.72 g / 12 g/mol = 2.48 moles

6.23 g/ 1 g/mol = 6.23 moles

Next, we divide the moles, by the lowest value of them (0.309)

0.309 / 0.309 = 1 mol Pb

2.48 / 0.309 = 8 mol C

6.23 / 0.309 = 20 mol H

There, we have our formula PbC₈H₂₀

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<em>n</em> = 15. A Bohr orbit with <em>n</em> = 15 comes closest to having a 24 nm diameter .

The formula for the radius <em>r</em> of the <em>n</em>th orbital of a hydrogen atom is

<em>r</em> = <em>n</em>^2·<em>a</em>

where

<em>a</em> = the Bohr radius = 0.0529 nm

We can solve this equation to get

<em>n</em> = √ (<em>r</em>/<em>a</em>)

If <em>d</em> = 24 nm, <em>r</em> = 12 nm.

∴ <em>n</em> = √(12 nm/0.0529 nm) = √227 = 15.1

<em>n</em> must be an integer, so <em>n</em> = 15.

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What is the maximum mass of ammonia that can be formed when 36.52 grams of nitrogen gas reacts with 10.62 grams of hydrogen gas
k0ka [10]

The maximum mass of NH₃ that can be formed when 36.52 g of N₂ reacts with 10.62 g of H₂ is 44.35 g

<h3>Balanced equation </h3>

N₂ + 3H₂ —> 2NH₃

Molar mass of N₂ = 14 × 2 = 28 g/mol

Mass of N₂ from the balanced equation = 1 × 28 = 28 g

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ from the balanced equation = 3 × 2 = 6 g

Molar mass of NH₃ = 14 + (3×1) = 17 g/mol

Mass of NH₃ from the balanced equation = 2 × 17 = 34 g

SUMMARY

From the balanced equation above,

28 g of N₂ reacted with 6 g of H₂ to produce 34 g of NH₃

<h3>How to determine the limiting reactant </h3>

From the balanced equation above,

28 g of N₂ reacted with 6 g of H₂

Therefore,

36.52 g of N₂ will react with = (36.52 × 6) / 28 = 7.83 g of H₂

From the above calculation, we can see that only 7.83 g out of 10.62 g of H₂ are required to react completely with 36.52 g of N₂.

Therefore, N₂ is the limiting reactant

<h3>How to determine the maximum mass of NH₃ produced </h3>

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃

Therefore,

36.52 g of N₂ will react to produce = (36.52 × 34) / 28 = 44.35 g of NH₃

Thus, the maximum mass of NH₃ obtained from the reaction is 44.35 g

Learn more about stoichiometry:

brainly.com/question/14735801

#SPJ1

4 0
2 years ago
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