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melomori [17]
1 year ago
8

Pop is made up of three major parts (water, carbon dioxide, and sugar). Assume that all other ingredients are present in insigni

ficant amounts. Explain how could you determine the moles and particles of water, carbon dioxide, and sugar in your pop. Be sure to show any necessary formulas in your answer.
Chemistry
1 answer:
Lady_Fox [76]1 year ago
7 0

If you were given the the total # of grams of all three compounds with the % of each molecular compound - H2O, CO2 and sugar [you would need to know the type of sugar in the Pop such as C12H22O11 as to find its MM (molar mass)]

To begin take the % of each compound x the total grams of all the compounds. For illustrative purposes let's say it works out to be 50 grams of H2O in the POP. The same would be. done for CO2 and for the sugar.

Step 1) With the mass of each you could determine the # of moles of each:

Example if the number grams of water in the sample is 50g, to determine the # of moles of water you would do the following - 50g H20 x 1 mol/18g H20 = 2.8 mol H2O The same technique would be used for the other compounds to find the # of moles.

Step 2) To find the representative particles of each(molecules, atoms) you would do the following:

as the example given of above for H20 - 50 grams you calculated as shown above to be the number of mol of H20 = 2.8mol

From the number of mol of H20, to determine the # of molecules of water you would set up the following:

2.8 mol H20 x 6.02 x 10^23/1 mol H2O = 1.69 X 10^24 molecules of H20.

The same would be done for CO2 and the sugar.

Step 3) Now to find the number of atoms of element of the compound taking for example the H2O example above:

Take the # of molecules of H2O found above and set it up in the following manner:

1.69 X 10^24 molecules H2O x 2 atoms H/1 molecule H2O = 3.38 x 10^24 atoms H

1.69 x 10^24 molecules H2O x 1 atom O)/1 molecule H2O = 1.69 X 10^24 atoms O

The same would be done for CO2 and for the sugar compound.

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<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<u>Explanation:</u>

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

<u>For lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol

  • The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = \frac{2}{1}\times 0.0021=0.0042mol of potassium bromide

  • Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g

  • To calculate the percentage composition of KBr in the mixture, we use the equation:

\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%

Hence, the percent by mass of KBr in the mixture is 9.996 %

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