Dear chem_tr
Thank You Very much for Your answer.
Here are my calculations.
First the chemical-reactions:
<br />
P_4 + 5O_2 \rightarrow \mathrm{P_4 O_{10}} \mathrm{(i)} <br />
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}
<br />
\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}<br />
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}
<br />
\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}<br />
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
