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10 months ago

A chemical reaction is shown below:

Chemistry

1 answer:

Vladimir [108]10 months ago

6 0

Answer:

Mass = 8.46 g

Explanation:

Given data Mass of water produced = ? Mass of glucose = 20 g Mass of oxygen = 15 g

Solution:

Chemical equation:

C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂

Number of moles of glucose:

Number of moles = mass/molar mass

Number of moles = 20 g/ 180.16 g/mol

Number of moles = 0.11 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 15 g/ 32 g/mol

Number of moles = 0.47 mol

now we will compare the moles of water with oxygen and glucose.

C₆H₁₂O₆ : H₂O

1 : 6

0.11 : 6/1×0.11 = 0.66

O₂ : H₂O

6 : 6

0.47 : 0.47

Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 0.47 mol ×18 g/mol

Mass = 8.46 g

Explanation:

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In a chemical equation the sum of the masses

monitta

The sum of the masses of the reactants must equal the sum of the masses of the products; as required by
the Principle of Conservation of Mass.

5 0

3 years ago

Calculate the feed ratio of adipic acid and hexamethylene diamine that should be employed to obtain a polyamide of approximately

artcher [175]

Answer:

r= 0.9949 (For 15,000)

r=0.995 (For 19,000)

Explanation:

We know that

Molecular weight of hexamethylene diamine = 116.21 g/mol

Molecular weight of adipic acid = 146.14 g/mol

Molecular weight of water = 18.016 g/mol

As we know that when adipic acid and hexamethylene diamine react then nylon 6, 6 comes out as the final product and release 2 molecule of water.

$M_{repeat}=146.14+166.21-2\times 18.106\ g/mol$

$M_{repeat}=226.32\ g/mol$

Mo= 226.32/2 =113.16 g/mol

$M_n=X_nM_o$

Given that

Mn= 15,000 g/mol

15,000 = Xn x 113.16

Xn = 132.55

Now by using Carothers equation we know that

$X_n=\dfrac{1+r}{1+r-2rp}$

$132.55=\dfrac{1+r}{1+r-2\times 0.99r}$

By calculating we get

r= 0.9949

For 19,000

19,000 = Xn x 113.16

Xn = 167.99

By calculating in same process given above we get

r=0.995

3 0

3 years ago

There is a photon with a frequency of 8x10^12 Hz. What is the energy of this<br> photon?

jarptica [38.1K]

Answer:

33.085476 meV ≈ 33 meV

Explanation:

Hope this helped!

7 0

3 years ago

aseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.5 g of methane is

maksim [4K]

Answer:

There is 9.6 grams of CO2 produced

Explanation:

Step 1: Data given

Mass of methane = 5.50 grams

Molar mass of methane = 16.04 g/mol

Mass of oxygen = 13.9 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The reaction

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles methane = 5.50 grams / 16.04 g/mol

Moles methane = 0.343 moles

Moles oxygen = 13.9 grams / 32.0 g/mol

Moles oxygen = 0.434 moles

For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

O2 is the limiting reactant. It will completely react (0.434 moles).

There will react 0.434/2 = 0.217 moles CH4

There will remain 0.343-0.217 = 0.126 moles CH4

There will be produced 0.434 moles of H2O and

0.434/2 =0.217 moles of CO2

Step 4: Calculate mass of products

Mass = moles * molar mass

Mass CO2 = 0.217 moles ¨44.01 g/mol

Mass CO2 = 9.6 grams

Mass H2O = 0.434 moles * 18.02

Mass H2O = 7.8 grams

4 0

2 years ago

Which gas will effuse at the rate closest At a particular pressure and temperature, nitrogen gas effuses at the rate of 79mLs. U

Contact [7]

Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$ ..........(1)

where,

$R_1$ = rate of effusion of nitrogen gas = $79mL/s$

$R_2$ = rate of effusion of sulfur dioxide gas = ?

$M_1$ = molar mass of nitrogen gas = 28 g/mole

$M_2$ = molar mass of sulfur dioxide gas = 64 g/mole

Now put all the given values in the above formula 1, we get:

$(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}$

$R_2=52mL/s$

Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.

4 0

3 years ago