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Ahat [919]
3 years ago
12

a student is given two 10g samples, each a mixture of only NaCL(s) and KCI(s) but im different proportions. how does knowing the

mass of chloride in each mixture allow you to determine which mixture has the higher proportion of KCI
Chemistry
1 answer:
shepuryov [24]3 years ago
3 0

Answer:

Knowing the mass of chlorine in each mixture will enable you find the empirical formulas that will lead to molecular formulas for the compounds in the mixture.This information will help you determine which mixture has a higher proportion of the compound in question.

Explanation:

In determining the percent composition of a compound in a mixture,mass of a compound will be the first step to be determined.This will then allow you to find mole to mole ratio of all elements in that compound.Having mole ratio information, and mass, it will be possible to find ratio of ions in the compound,and determine the mixture with higher proportion of the compound.

Molecular weight and the empirical formula of the compound will led you to the molecular formula of the compound,that will show you the mole ratios which now you can use to determine proportions of the compound in a mixture.

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If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically
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548.55 grams of aluminum hydroxide should theoretically form.

Explanation:

Balanced equation for the reaction:

3 NaOH + Al ⇒ Al(OH)3 +3 Na

DATA GIVEN:

mass of NaOH = 842 grams, atomic mass =39.9 grams/mole

mass of Al = 750 grams, atomic mass = 26.9 grams/mole

aluminum hydroxide theoretical yield = ?

Moles of NaOH reacted

number of moles = \frac{mass}{atomic mass of 1 mole}

putting the values in the equation

NaOH = \frac{842}{39.9}

           = 21.1 MOLES OF NaOH

Al = \frac{750}{26.9}

   = 27.8 moles

from the equation

 from 3 moles of NaOH 1 mole of Al(OH)3 is produced

21.1 moles of NaOH will react to give x moles of Al(OH)3

\frac{1}{3} = \frac{x}{21.1}

7.03 moles of Al(OH)3 is formed.

and

1 mole of Al(OH)3 is formed from 1 mole of Al in the reaction

so, 27.8 Moles will react to give give 27.8 moles of Al(OH)3 limiting reagent of the given reaction is NaOH

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theoretical  yield from the given data is 548.55 grams

3 0
3 years ago
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