Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Questions regarding Chemical Reactions

  1. Oct 5, 2004 #1

    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.

  2. jcsd
  3. Oct 5, 2004 #2


    User Avatar
    Science Advisor
    Gold Member

    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:

    P_4 + 5O_2 \rightarrow \mathrm{P_4 O_{10}} \mathrm{(i)}

    [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]

    \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}

    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]

    \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}

    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.

  5. Oct 5, 2004 #4


    User Avatar
    Science Advisor
    Gold Member


    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


    User Avatar

    Staff: Mentor

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Alex72, please start a new thread with your polymer reaction question.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook