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Questions regarding Chemical Reactions

  1. Oct 5, 2004 #1
    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
     
  2. jcsd
  3. Oct 5, 2004 #2

    chem_tr

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    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
     
  4. Oct 5, 2004 #3
    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
     
  5. Oct 5, 2004 #4

    chem_tr

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    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.
     
  6. Jun 12, 2009 #5
    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.....?
     
  7. Jun 12, 2009 #6

    Borek

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    This thread was active five years ago...
     
  8. Jun 12, 2009 #7

    chemisttree

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    Alex72, please start a new thread with your polymer reaction question.
     
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