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 I've 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.