What weights of $P_{4} O_{6} a n d P_{4} O_{10}$ will be produced by the combustion of $o f 31 g o f P_{4} i n 32 g$ of oxygen leaving $n o P_{4} a n d O_{2}$ ?

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$27.5 g, 35.5 g$

$55 g, 71 g$

$2.75 g, 219.5 g$

$17.5 g, 190.5 g$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$Percentage of P_{4} O_{6} = \frac{(31 \times 4) \times 100}{(31 \times 4 + 32 \times 6)} = 56.3437 % P$

$Percentage of P_{4} O_{10} = {43.6420}^{\circ} P$

$31 gP = 31 / (31 + 32) = 49.2063 % P$

So,

$\begin{aligned} 100 % P_{4} O_{6} = 56.3437 % \\ 2 % P_{4} O_{6} = 49.2063 % \\ 0 % P_{4} O_{6} = 43.6420 % \end{aligned}$

Find the value of z

$\begin{aligned} \frac{z}{100} = (49.2063 - 43.6420) (56.3437 - 43.6420) \\ z = 43.808 % P_{4} O_{6} \end{aligned}$

By the law of the Conservation of Matter, the total mass of $P_{4} O_{6} and P_{4} O_{10} = 31 g + 32 g = 63 g$

$43.808 % of 63 g = 27.6 {gP}_{4} O_{6}$

$63 g - 27.6 g = 35.4 {gP}_{4} O_{10}$

Watch 3-min video & get full concept clarity