Questions regarding Chemical Reactions

[Mathman23]

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

[chem_tr]

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

[Mathman23]

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)}


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

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

\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}


\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 H_3 PO_{4} then the mass of P_4 is 35 grams.

\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}


\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 O_{2} used in (ii)

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

n({O_2}) = 1,41 \mathrm{mol}

R = 0,083 \mathrm{\frac{L \cdot bar}{mol \cdot K}}

then the volume 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}

Here is there I have a problem:

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

Thank You very much again for Your kind answer.

Sincerely
Fred

[chem_tr]

Hello,

110,5 grams of H_3PO_4 is 1,123 mol, as you found in your thread. As C= \frac {n}{V}, you may rearrange this equation to find V:
V= \frac {n}{C}= and this is up to you.

[alex72]

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.....?

[Borek]

This thread was active five years ago...

[chemisttree]

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