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sergij07 [2.7K]
2 years ago
15

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In

the first step, calcium carbide and water react to form acetylene and calcium hydroxide:CaC2(s)+2H2O(g)→C2H2(g)+CaOH2(s)In the second step, acetylene, carbon dioxide and water react to form acrylic acid:6C2H2(g)+3CO2(g)+4H2O(g)→5CH2CHCO2H(g)Write the net chemical equation for the production of acrylic acid from calcium carbide, water and carbon dioxide. Be sure your equation is balanced.
Chemistry
2 answers:
svet-max [94.6K]2 years ago
8 0

Answer:

6 CaC₂(s) + 16 H₂O(g) + 3 CO₂(g) → 6 Ca(OH)₂(s) + 5 CH₂CHCO₂H(g)

Explanation:

There are two steps in the preparation of acrylic acid.

Step 1: CaC₂(s) + 2 H₂O(g) → C₂H₂(g) + Ca(OH)₂(s)

Step 2: 6 C₂H₂(g) + 3 CO₂(g) + 4 H₂O(g) → 5 CH₂CHCO₂H(g)

In the net chemical equation, intermediaries do not appear, C₂H₂ in this case. To achieve this, we will multiply the first step by 6, add it to the second step and cancel what is on both sides of the equation.

6 CaC₂(s) + 12 H₂O(g) → 6 C₂H₂(g) +  6Ca(OH)₂(s)

+

6 C₂H₂(g) + 3 CO₂(g) + 4 H₂O(g) → 5 CH₂CHCO₂H(g)

-----------------------------------------------------------------------------------------------------

6 CaC₂(s) + 12 H₂O(g) + 6 C₂H₂(g) + 3 CO₂(g) + 4 H₂O(g) → 6 C₂H₂(g) +  6Ca(OH)₂(s) + 5 CH₂CHCO₂H(g)

6 CaC₂(s) + 16 H₂O(g) + 3 CO₂(g) → 6 Ca(OH)₂(s) + 5 CH₂CHCO₂H(g)

Oliga [24]2 years ago
7 0

You add the both equations, remove the molecules that may be found in the same time at the right and left of the reaction and balance the equation. Doing so you will get:

6 CaC₂ (s) + 16 H₂O (g) + 3 CO₂ (g) → 5 CH₂=CH-COOH (g) + 6 Ca(OH)₂ (s)

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How many liters of 3.0 M NaOH solution will react with 2.4 mol H2SO4? (Remember to balance the equation.)
timurjin [86]

Answer:

1.6 L is the volume of NaOH that has reacted

Explanation:

The balanced reaction is:

H₂SO₄ + 2NaOH → Na₂SO₄  + 2H₂O

This is a neutralization reaction between a strong acid and a strong base. The products are the correspond salt and water.

We propose this rule of three:

1 mol of sulfuric acid needs 2 mol of NaOH to react to react

Then, 2.4 moles of H₂SO₄  will react with (2.4 . 2) / 1 = 4.8 moles of NaOH

As molarity is 3M, we can determine the volume of our solution

Molarity (M) = mol / volume(L) → Volume(L) = mol / Molarity

Volume(L) = 4.8 mol / 3 M = 1.6 L

3 0
2 years ago
Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W
nikklg [1K]

Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

     H2= (1-x)Haq+XHvap.........1

    Putting the values in 1

    2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

                        = 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

                     x = 1904.976 (J/g)/1998.576 (J/g)

                     x = 0.953

So, the quality of the wet steam is 0.953

                   

7 0
2 years ago
Read 2 more answers
If a concentration of 10 fluorescent molecules per μm2of cell membrane is needed to visualize a cell under the microscope, how m
lara [203]

Answer : Total molecules that will be needed to visualize a single egg will be 78500 molecules of dye.

Explanation : As a single egg cell has an approximately diameter of 100 μm.

We can use this formula to calculate area of the cell membrane;

A = π (100)^{2} / 4;  

We can take π as 3.14 and we get;

A = 3.14 X (100)^{2} / 4  

Soving we get;

A =  7850 μm^{2}  

Here we have to calculate the amount of dye molecules which will be needed for 10 fluorescent molecules / μm^{2}  but;

here 1 μm^{2} = 7850 μm^{2} dye molecules.

Therefore, 10 fluorescent molecules will need;  

7850 X 10 = 78500 molecules of dye.

Therefore, the answer is 78500 molecules of dye.

6 0
2 years ago
How many lead (Pb) atoms will be generated when 5.38 moles of ammonia react according to the following equation: 3PbO+2NH3→3Pb+N
jasenka [17]

Answer:

4.86×10^23 molecule of Pb

Explanation:

Based on that equation, for every 2 moles of ammonia, you get 3 moles of lead.

So:

2 mol NH3/ 3 mol Pb

Using this ratio we can find the amounts of either molecule. Given 5.38 mol NH3:

(5.38 NH3)(3 Pb/ 2 NH3) = (5.38)(3/2) mol Pb = 8.07 mol Pb

Then, we just need to use Avagadro's number to get the number of molecules.

(8.07)(6.02×10^23) = 4.86×10^23 molecule of Pb

4 0
2 years ago
Read 2 more answers
The chlorination of methane occurs in a number of steps that results in the formation of chloromethane and hydrogen chloride. Th
kenny6666 [7]

Answer:

Total pressure = 0,806 atm

Partial pressure of CH₄: 0,037 atm

Partial pressure of Cl₂: 0,396 atm

Partial pressure of CH₃Cl: 0,125 atm

Partial pressure of HCl: 0,125 atm

Partial pressure of Cl⁻: 0,125 atm

Explanation:

For the reaction:

2CH₄(g)+3Cl₂(g)⟶2CH₃Cl(g)+2HCl(g)+2Cl⁻(g)

295 mL≡ 0,295L of methane at STP are:

n = PV/RT

P = 1 atm; V = 0,295L; R = 0,082atmL/molK; T = 273K.

moles of methane: 0,0132 moles

For 725 mL of chlorine ≡ 0,725L

moles of chlorine at STP are: ≡ 0,0324 moles

For a complete reaction of 0,0132 moles of CH₄:

0,0132 mol CH₄× \frac{3molCl_{2}}{2 molCH_{4}} = <em>0,0198 moles</em>

The reaction reaches 77%, moles of Cl₂ that react are: 0,0198×77% = 0,0153 mol

As you have 0,0324 moles of Cl₂, moles that will not react are:

0,0324 - 0,0153 = <em>0,0171 mol Cl₂</em>

As the reaction reaches 77% completion, moles of CH₄ that react are:

0,0132×77% =<em> 0,0102 moles of CH₄ And the moles that don't react are </em><em>0,00300 mol</em>

Thus, moles of each compound are:

0,0102 moles of CH₄×\frac{3Cl_{2}}{2 molCH_{4}}= <em>0,0153 mol  + 0,0171 mol = 0,0324 mol Cl₂</em>

0,0102 moles of CH₄×\frac{2CH_{3}Cl}{2 molCH_{4}}= <em>0,0102 mol CH₄</em>

0,0102 moles of CH₄×\frac{2HCl}{2 molCH_{4}}= <em>0,0102 mol HCl</em>

0,0102 moles of CH₄×\frac{2Cl^{-}}{2 molCH_{4}}= <em>0,0102 mol Cl⁻</em>

Total pressure using:

P = nRT/V

Where: n = 0,0102mol×3+0,0324mol + 0,0030mol = 0,0660mol; R = 0,082 atmL/molK; T = 298K; V = 2L

Total pressure = 0,806 atm

Partial pressure of CH₄: 0,037 atm

Partial pressure of Cl₂: 0,396 atm

Partial pressure of CH₃Cl: 0,125 atm

Partial pressure of HCl: 0,125 atm

Partial pressure of Cl⁻: 0,125 atm

<em>-To obtain partial pressure you change the moles for each compound-</em>

<em />

I hope it helps!

5 0
2 years ago
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