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3 years ago

Select the correct answer.

Chemistry

1 answer:

Phoenix [80]3 years ago

7 0

Answer:

Molecular formula = C₆H₁₂O₆

Explanation:

Given data:

Mass of hydrogen = 31.7 g

Mass of carbon = 283.4 g

Mass of oxygen = 377.4 g

Molar mass of compound = 176.124 g/mol

Molecular formula = ?

Solution:

Number of gram atoms of H = 31.7 / 1.01 = 31.4

Number of gram atoms of O = 377.4 / 16 = 23.6

Number of gram atoms of C = 283.4 / 12 = 23.6

Atomic ratio:

C : H : O

23.6/23.6 : 31.4/23.6 : 23.6/23.6

1 : 1.33 : 1

C : H : O =3 (1 : 1.33 : 1 )

Empirical formula is C₃H₄O₃.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = 3×12+4+3×16 = 88

n = 176.124 / 88

n = 12

Molecular formula = n (empirical formula)

Molecular formula = 2 (C₃H₄O₃)

Molecular formula = C₆H₁₂O₆

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A mixture 21.2 g of P and 79.0 g Cl2 reacts completely to form PCl3 and PCl5 as the only products. find the mass of PCl3

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Considering the definition of reaction stoichiometry and limiting reagent, the mass of PCl₃ that could be formed is 38.26 grams.

In first place, you must know that the balanced reaction is:

4 Cl₂ + 2 P → PCl₃ + PCl₅

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Cl₂: 4 moles
P: 2 moles
PCl₃: 1 mole
PCl₅: 1 mole

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Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

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PCl₅: 1mole× 208.25 g/mole= 208.25 grams

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 62 grams of P reacts with 283.6 grams of Cl₂, 21.2 grams of P reacts with how much mass of Cl₂?

$mass of Cl_{2}=\frac{21.2 grams of Px283.6 grams of Cl_{2} }{62 grams of P}$

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But 96.97 grams of Cl₂ are not available, 79 grams are available. Since you have less mass than you need to react with 21.2 grams of P, Cl₂ will be the limiting reagent.

Then, it is possible to determine the mass of PCl₃ produced by another rule of three, using the limiting reagent: if by stoichiometry 283.6 grams of Cl₂ produce 137.35 grams of PCl₃, 79 grams of Cl₂ how much mass of PCl₃ will be formed?

$mass of PCl_{3}=\frac{137.35 grams of PCl_{3}x79 grams of Cl_{2} }{283.6grams of Cl_{2}}$

mass of PCl₃= 38.26 grams

In summary, the mass of PCl₃ that could be formed is 38.26 grams.

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