Questions regarding Chemical Reactions

  • Thread starter Mathman23
  • Start date
  • #1
254
0
Hi

I got two questions regarding the following chemical:

i) P_4 + 5O_2 -----> P_4 O_10

ii) P_4 O_10 + 6H_2O -----> 4H3 PO_4

How many grams of P_4 O_10 are formed in i) if there is 35 grams of P_4 ??

How many grams of H_3 PO_10 are formed in ii) if I have 35 grams of P_4 ??

Thanks in advance.

Sincerely
Fred
 

Answers and Replies

  • #2
chem_tr
Science Advisor
Gold Member
609
3
Dear Fred,

P4 with every P of 31 g/mol makes 124 grams for this chemical (P:31, O:16, H:1 gram/mol). Find the appropriate mole number from there. Remember that 1:4 mole ratio is present to convert this into H3PO4.

I am sure you can find the grams of product as the reaction proceeds 100%.

Regards, chem_tr
 
  • #3
254
0
Dear chem_tr

Thank You Very much for Your answer.

Here are my calculations.

First the chemical-reactions:

[tex]
P_4 + 5O_2 \rightarrow \mathrm{P_4 O_{10}} \mathrm{(i)}
[/tex]

[tex] P_4 O_{10} + 6 H_2 O \rightarrow 4 H_3 PO_{4} \mathrm{(ii)} [/tex]

a) Calculating the mass of [tex] \mathrm{P_{4} O_{10}} [/tex] then the mass of [tex] \mathrm{P_4} [/tex] is 35 grams.

[tex] \frac{n(P_4 O_{10})}{n(P_{4})} = 1 \rightarrow n(P_{4} O_{10}) = n(P_{4}) = \frac{m(P_4)}{M(P_4)} = \frac{35,00 \mathrm{g}}{124,00 \mathrm{g/mol}} = 0,282 \mathrm{mol} [/tex]

[tex]
\mathrm{m(P_4 O_{10})} = \mathrm{M(P_4 O_{10})} \cdot n(P_4 O_{10}) = 284,00 \mathrm{g/mol} \cdot 0,282 \mathrm{mol} = 80,09 \mathrm{g}
[/tex]

b) Calculating the mass of [tex] H_3 PO_{4} [/tex] then the mass of [tex] P_4 [/tex] is 35 grams.

[tex] \frac{n(H_3 PO_{4})}{n(P_{4} O_{10})} = 4 \rightarrow n(P_{4} O_{10}) = 4 \cdot n(H_3 PO_{4}) = 4 \cdot \frac{80,09 \mathrm{g}}{284,00 \mathrm{g/mol}} = 4 \cdot 0,282 \mathrm{mol} = 1,123 \mathrm{mol}[/tex]

[tex]
\mathrm{m(H_3 PO_{4})} = \mathrm{M(H_3 PO_{4})} \cdot \mathrm{n(H_3 PO_{4})} = 98,00 \mathrm{g/mol} \cdot 1,123 \mathrm{mol} = 110,5 \mathrm{g}
[/tex]

c) The Volume of [tex] O_{2} [/tex] used in (ii)

P = 1,0 bar
T = 25 + 273 = 298 K

[tex] n({O_2}) = 1,41 \mathrm{mol} [/tex]

[tex] R = 0,083 \mathrm{\frac{L \cdot bar}{mol \cdot K}} [/tex]

then the volume [tex] V = \frac{1,41 mol \cdot 0,083 \mathrm{\frac{L \cdot bar}{mol \cdot K}} \cdot 298 K }{1,0 \mathrm{bar}} = 34,9 \mathrm{L} [/tex]

Here is there I have a problem:

d) How many liters of 0,500 M [tex]H_3 PO_{4}[/tex] can be generated by the [tex]H_3 PO_{4}[/tex] in b ?

Thank You very much again for Your kind answer.

Sincerely
Fred
 
  • #4
chem_tr
Science Advisor
Gold Member
609
3
Hello,

110,5 grams of [itex]H_3PO_4[/itex] is 1,123 mol, as you found in your thread. As [itex]C= \frac {n}{V}[/itex], you may rearrange this equation to find V:
[tex]V= \frac {n}{C}= [/tex] and this is up to you.
 
  • #5
3
0
Dear Fred,

P4 with every P of 31 g/mol makes 124 grams for this chemical (P:31, O:16, H:1 gram/mol). Find the appropriate mole number from there. Remember that 1:4 mole ratio is present to convert this into H3PO4.

I am sure you can find the grams of product as the reaction proceeds 100%.

Regards, chem_tr

dear chem_tr, please sxcuse me jumping in here ive just joined and i need to ask a general Q re polymer reactions. should i continue or.....?
 
  • #6
Borek
Mentor
28,888
3,432
This thread was active five years ago...
 
  • #7
chemisttree
Science Advisor
Homework Helper
Gold Member
3,678
679
Alex72, please start a new thread with your polymer reaction question.
 

Related Threads on Questions regarding Chemical Reactions

Replies
2
Views
4K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
22
Views
4K
Top