Question

Consider the chemical equations shown here.

P4(s) + 302(g) – P406(s)
P4(s) + 502(g) – P4010(s)
What is the overall equation for the reaction that produces P4010 from P406 and O2?
o P40g(s) + O2(g) → P4010(s)
o P406(s) + 202(9) – P4010(5)
0 P406(S) + 802(g) – P4010(5)

Answer 1

B. P406(s) + 202(g) –> P4010(s)

Step-by-step explanation:

got it right on edge

Answer 2

Step-by-step explanation:

The overall equation for the reaction that produces P4010 from P406 and O2 is:

P4O6(s) + 8O2(g) → P4O10(s)

This equation can be obtained by adding the two given equations and canceling out the common reactants and products:

P4(s) + 3O2(g) → P4O6(s)

P4(s) + 5O2(g) → P4O10(s)

By multiplying the first equation by 4 and the second equation by 2, we can get the same number of phosphorus atoms on both sides:

4P4(s) + 12O2(g) → 4P4O6(s)

2P4(s) + 10O2(g) → 2P4O10(s)

Adding these two equations gives:

6P4(s) + 22O2(g) → 4P4O10(s) + 2P4O6(s)

Since P4O6 is not the desired product, we can eliminate it from the equation by subtracting 2P4O6(s) from both sides:

6P4(s) + 22O2(g) → 4P4O10(s)

This is the same as:

P4O6(s) + 8O2(g) → P4O10(s)

Therefore, the overall equation for the reaction that produces P4010 from P406 and O2 is:

P4O6(s) + 8O2(g) → P4O10(s