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3 years ago

Calculate the mass percent by volume of 330.1 g of glucose (C₆H₁₂O₆, MM = 180.2 g/mol) in 325 mL of solution.

Chemistry

1 answer:

nadezda [96]3 years ago

3 0

Answer: The mass percent by volume is 101.6%

Explanation:

The solution concentration expressed in percent by volume means that the amount of solute present in 100 parts volume of solution.

It is represented in formula as :

mass percent by volume = $\frac{\text {mass of solute}\times 100}{\text {Volume of solution in ml}}\%$

Given : mass of glucose = 330.1 g

volume of solution = 325 ml

Thus mass percent by volume = $\frac{330.1g\times 100}{325ml}=101.6\%$

Thus the mass percent by volume is 101.6%

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Answer is: 2. atomic number.

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3 years ago

Gaseous methane (CH4) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H20). What

Mariana [72]

Answer:

Theoretical yield = 3.51 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $CH_4$ :-

Mass of $CH_4$ = 1.28 g

Molar mass of $CH_4$ = 16.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.28\ g}{16.04\ g/mol}$

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For $O_2$ :-

Mass of $O_2$ = 10.1 g

Molar mass of $O_2$ = 31.998 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{10.1\ g}{31.998\ g/mol}$

$Moles_{O_2}= 0.3156\ mol$

According to the given reaction:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

1 mole of methane gas reacts with 2 moles of oxygen gas

0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas

Moles of oxygen gas = 0.1596 moles

Available moles of oxygen gas = 0.3156 moles

Limiting reagent is the one which is present in small amount. Thus, $CH_4$ is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction produces 1 mole of carbon dioxide.

0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.

Mole of carbon dioxide = 0.0798 mole

Molar mass of carbon dioxide = 44.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0798\ moles= \frac{Mass}{44.01\ g/mol}$

Mass of $CO_2$ = 3.51 g

Theoretical yield = 3.51 g

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