Question

What weights of $P_4{O}_6$ and $P_4{O}_{10}$ will be produced by the combustion of 31 g of $P_4$ with 32g of oxygen leaving no $P_4$ and $O_2$?

• A. 2.75 g, 219.5 g
• B. 27.5 g, 35.5 g
• C. 55 g, 71 g
• D. 17.5 g, 190.5 g

Answer 1

Answer: (B)

27.5 g, 35.5 g

Percentage of $P_4{O}_6=\frac{(31\times 4)\times 100}{(31\times 4+32\times 6)}=56.3437\%P$

Percentage of $P_4{O}_{10}=43.6420\%P$

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

So,

100% $P_4{O}_6$ = 56.3437%

z% $P_4{O}_6$ = 49.2063%

0% $P_4{O}_6$ = 43.6420%

Find the value of z

$\frac{z}{100}=(49.2063-43.6420)/(56.3437-43.6420)$

$z=43.808$% $P_4{O}_6$

By the Law of the Conservation of Matter, the total mass

of $P_4{O}_6$ and $P_4{O}_{10}$ = 31g + 32g = 63g

43.808% of 63g = 27.6 g $P_4{O}_6$

$63g-27.6g=35.4g$ $P_4{O}_{10}$