Correct Answer: Option B

Explanation

P\(_4\)  + 50\(_2\) → P\(_4\)O\(_{10}\)

for the complete combustion of 124 grammes of phosphorus we need

5 x 22.4 litres of 0\(_2\)

for 12.4 gms of P\(_4\)  we need

\(\frac{5 \times 22.4}{124} \times \frac{12.4}{1}\)

\(\frac{22.4}{2}\) = 11.2 litres

Alternatively, 

                            P\(_4\)    +    5 0\(_2\)  →  P\(_4\)O\(_{10}\)

                            1 mole  :   5 moles

 Mass of P\(_4\)  = 12.4

Molar mass of P\(_4\)  = (31 x 4)=124g/mol

                    n = \(\frac{mass}{Molar mass}\)

                       = \(\frac{12.4}{124}\)

                       = 0.1 mole

Then               0.1 mole P\(_4\)  will react with   0.5 mole 0\(_2\)

But                                                              n = \(\frac{Volume}{Molar Volume}\)

                                                                  0.5 = \(\frac{Volume}{22.4}\)

                                                                   V = 0.5 X 22.4

                                                                    V = 11.20 Litres

• if 5^(2x)=125 find x? Answer this

  3 Answers   ·   Mathematics   ·   4 days ago
• how does post utme looks like and what are the subjects required for a medical student? Answer this

  2 Answers   ·   2 days ago
• What are the male and female reproductive hormones in animals? Answer this

  4 Answers   ·   Agricultural Science   ·   2 days ago