Question

Consider a situation in which 186 g of P4 are exposed to 208 g of O2.

a) What is the maximum amount in moles of P2O5 that can theoretically be made from 186 g of P4 and excess oxygen?

b) What is the maximum amount in moles of P2O5 that can theoretically be made from 208 g of O2 and excess phosphorus?

How do I solve these? I forgot the equations

Thanks in advance

Answer 1

Answer:  (a) 3.00 moles P2O5

(b) 2.6 moles P2O5.

Explanation :

the first step to solve this problem is to write down the related reaction and to balance them. Which is written as-

P4(s) + O2(g) ------------> P2O5(s)

First balance phosphorus, there is 4 P at Left hand side of the equation while at right hand side only 2 P is present thus to balance them ,we multiply P2O5 by 2.

P4(s) + O2(g) -----------> 2P2O5(s).

Now , we balance the oxygen atom, two oxygen is Left hand side of the equation while at right hand side of the equation ,Now , 10 oxygen is present to balance the equation, we multiply O2 by 5.

So, the balance equation is :

P4(s) + 5 O2(g) ---------> 2 P2O5(s).

In the next step , We have to convert the mass of P4 in moles of P4. And again , into the moles of P2O5.

This is done by the following formula,

Since we have to note down the molar mass of P4 = 123.90 g ,

Now,

Maximum amount in moles of P2O5 =186 g P4 * 1 mole P4/ 123.90 g P4 * 2 mole P2O5 /1 mole P4.

= 3.00 moles P2O5.

Answer part b)

The maximum amount in moles of P2O5 that can theoretically be made from 208 g of O2 and excess phosphorus is calculated by this formula.

= (208 g O2)* 1 mole O2/ 32.00 g O2. * 2 moles P2O5/5 moles O2.

= 416/160 = 2.6 moles P2O5.

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