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hoa [83]
2 years ago
7

A sample of a certain binary compound contains 6.93 g of silicon and 7.89 g of oxygen. What is the percent composition of the co

mpound?
Chemistry
2 answers:
sp2606 [1]2 years ago
5 0

Given, mass of the silicon = 6.93 g

mass of the oxygen = 7.89 g

Calculate the number of moles of both atoms:

Number of moles = \frac{given mass}{molar mass}

Insert the given values of mass in above formula:

Number of moles of silicon = \frac{given mass of silicon in g}{molar mass of silicon}

= \frac{6.93 g}{28 g/mol}

= 0.2475 mol

Number of moles of oxygen = \frac{given mass of silicon in g}{molar mass of silicon}

= \frac{7.89 g}{16 g/mol}

= 0.493125 mol

Thus,

Total number of moles =  number of silicon+ number of mole of oxygen

= 0.493125 mol+0.2475 mol =0.740625 mol

Percentage composition = \frac{number of moles of an atom}{total number of moles of the compound}\times 100

Put the values in above formula:

Percentage composition of silicon= \frac{number of moles of silicon}{total number of moles of the compound}\times 100

Percentage composition of silicon= \frac{0.2475 mol}{0.740625 mol}\times 100

= 33.41 percent

Percentage composition of oxygen= \frac{number of moles of oxygen}{total number of moles of the compound}\times 100

Percentage composition of silicon= \frac{0.493125 mol}{0.740625 mol}\times 100

= 66. 58 percent

Thus, percent composition of silicon and oxygen is 33.41 percentand 66. 58 percent

TiliK225 [7]2 years ago
4 0
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fredd [130]

Answer:

a. 510.6 g of C₂H₅OH are produced from 1kg of glucose

b. 171.1 g of glucose are required

Explanation:

Chemist reaction is this:

C₆H₁₂O₆  →  2C₂H₅OH(l) + 2CO₂(g)

So 1 mol of glucose can produce 1 mol of ethyl alcohol.

First of all, we should convert the mass to g, afterwards to moles

1 kg . 1000 g/ 1kg = 1000 g . 1 mol/180 g = 5.55 moles

Then we can think, this rule of three

1 mol of glucose can produce 2 moles of ethyl alcohol

Then 5.55 moles of glucose may produce the double of moles of C₂H₅OH

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Let's convert the moles to mass → 11.1 mol . 46g /1mol = 510.6 g

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1 gasohol is 10 mL C₂H₅OH / 90 ml of gasoline. We should make a rule of three.

In 90 mL of gasoline we have 10 mL of C₂H₅OH

In 1000 mL (1L) we would have (1000 . 10)/ 90 = 111.1 mL

Now we have to determine the mass of C₂H₅OH that is contained in the volume we have calculated. We must use the density.

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Now, we convert the mass to moles → 87.7 g . 1mol/ 46g = 1.91 mol

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A chemical waste label does not mention or describe the experiment for which the chemicals were used for, scientists, teacher or students should have knowledge of the chemical composition by reading their names and evaluate themselves on which chemical should be used for which experiment.

Hence, the correct answer is  "A description of the experiment the chemicals were used for."

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