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Gekata [30.6K]
3 years ago
6

Phosphoric acid, which is commonly used as rust inhibitor, food additive and etching agent for dental and orthopedic use, can be

synthesized using a two-step thermal process. in the first step, phosphorus and oxygen react to form diphosphorus pentoxide: (l) (g) (g)
Chemistry
1 answer:
igor_vitrenko [27]3 years ago
6 0

The first step in the synthesis of phosphoric acid is the reaction of phosphorus and oxygen to form diphosphorus pentoxide.

P_{4}(l) + 5 O_{2}(g)--->2 P_{2}O_{5} (g)

The second step in the synthesis of phosphoric acid is the reaction of diphosphorus with water to form phosphoric acid.

P_{2}O_{5}(g) + 3H_{2}O (l) --> 2H_{3}PO_{4}(aq)

The net chemical equation representing the formation of phosphoric acid from phosphorus, water and oxygen will be,

Step -1 : P_{4}(l) + 5O_{2}(g) ---> 2P_{2}O_{5}(g)

Step 2: P_{2}O_{5}(g) + 3 H_{2}O(l)-->2 H_{3}PO_{4}(aq)

Equation in step 2 must be multiplied by 2 in order to cancel diphosphorus pentoxide.

P_{4}(l) + 5O_{2}(g)---> 2P_{2}O_{5}

2 [P_{2}O_{5} (g) + 3 H_{2}O(l) --> 2 H_{3}PO_{4} (aq) ]

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

Net: P_{4}(l) + 5 O_{2}(g) + 6 H_{2}O(l) --->   4H_{3}PO_{4}(aq)


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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.

So

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

M_{repeat}=226.32\ g/mol

So

Mo= 226.32/2 =113.16 g/mol

M_n=X_nM_o

Given that

Mn= 15,000 g/mol

So

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
Consider the reaction 2 al + Fe2O3 to 2Fe + Al2O3. If 60.0g of Al is reacted with excess Fe2O3, determine the amount (in moles)
olganol [36]

 The  amount  of  Al2O3  in moles=  1.11 moles    while in  grams   = 113.22 grams


    <em><u>calculation</u></em>

     2 Al  + Fe2O3 → 2Fe  + Al2O3

    step  1: find the moles of Al  by  use of <u><em>moles= mass/molar  mass  </em></u>formula

    =  60.0/27= 2.22  moles


    Step 2: use the mole ratio to determine the  moles of Al2O3.

 The  mole ratio  of Al : Al2O3 is  2: 1 therefore the moles of Al2O3= 2.22/2=1.11  moles


Step 3:    finds the mass  of  Al2O3  by us of  <u><em>mass= moles x molar mass</em></u><em> </em>formula.

The molar  mass of Al2O3  =  (2x27)  +( 16 x3) = 102  g/mol

mass is therefore=  102  g/mol  x 1.11= 113.22 grams


             

7 0
3 years ago
nitrogen is made up of two isotopes N-14 and N15. given nitrogen’s atomic weight of 14.007 what is the percent abundance of each
Oduvanchick [21]
The answer is n14 if it
8 0
3 years ago
A sample of a compound is decomposed in the laboratory and produces 330 g carbon, 69.5 g hydrogen, and 220.2 g oxygen. Calculate
Zarrin [17]

Answer: Empirical formula is C_2H_5O

Explanation: We are given the masses of elements present in a sample of compound. To evaluate empirical formula, we will be following some steps.

<u>Step 1 :</u> Converting each of the given masses into their moles by dividing them by Molar masses.

Moles=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of Carbon = 12.0 g/mol

Molar mass of Hydrogen = 1.0 g/mol

Molar mass of Oxygen = 16.0 g/mol

Moles of Carbon = \frac{330g}{12g/mol}=27.5moles

Moles of Hydrogen = \frac{69.5g}{1g/mol}=69.5moles

Moles of Oxygen = \frac{220.2g}{16g/mol}=13.76moles

<u>Step 2: </u>Dividing each mole value by the smallest number of moles calculated above and rounding it off to the nearest whole number value

Smallest number of moles = 13.76 moles

\text{Mole ratio of Carbon}=\frac{27.5moles}{13.76moles}=1.99\approx 2

\text{Mole ratio of Hydrogen}=\frac{69.5moles}{13.76moles}=5.05\approx 5

\text{Mole ratio of Oxygen}=\frac{13.76moles}{13.76moles}=1

<u>Step 3:</u> Now, the moles ratio of the elements are represented by the subscripts in the empirical formula

Empirical formula becomes = C_2H_5O

7 0
3 years ago
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