Question

What is the final chemical equation from the following intermediate chemical equations?

P.O.(s) → P(s)+30,(9)
P.(s)+502(g) →P,01(s)
PaOo(s) + 2O2(g) → P4010(s)
P400(s) + 802(g) → 2P4(s) + P4010(s)
P4Oo(s) + 1502(g) → P4(s) + P4010(s)
P400(S)+(5/3)02(g) → PO10(s)

Answer 1

A

Step-by-step explanation:

the P4(s) s cancel out on both sides and the 3O2(g) is crossed out while the 5O2(g) on the left is turned into a 2O2(g) because the 3O2 is subtracted from it. SO it should leave you with A

Answer 2

The final chemical equation would be :

A.

`P_(4)O_(6)+2O_(2)\rightarrow P_(4)O_(10)`

Step-by-step explanation:

This is obtained by adding the respective intermediates:

`P_(4)O_(6)\rightarrow P_(4) + 3O_(2)`

`P_(4)+5O_(2)\rightarrow P_(4)O_(10)`

P4(s) are on opposite side of arrow so cancel each other.(No P4 in final reaction)

O2(g) = 5 - 3 = 2

There are 2 O2 more on left (reactant side)

finally there are ,

P4O6 and 2 O2 on left hand side and P4O10 on Right side of the arrow

So we get

`P_(4)O_(6)+2O_(2)\rightarrow P_(4)O_(10)`