Question

Part C

In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Part B, you found the amount of product (1.40 mol P2O5 ) formed from the given amount of oxygen and excess phosphorus.

Now, determine how many moles of P2O5 are produced from the given amounts of phosphorus and oxygen.

Answer 1

Chemical equation,

2P2 + 5O2 --> 2P2O5

moles of product formed = 1.80 mol in Part A.

moles of product formed = 1.40 mol in Part B

Calculation,

moles of phosphorous = grams of phosphorous/molar mass of phosphorous

moles mass of phosphorous = 62 g/mol

Oxygen is in excess, so limiting reactant will be phosphorous

moles of P2O5 formed from given phosphorous = moles of phosphorous reacted

So,

For 1.80 moles P2O5 in Part A, we had 1.80 moles of Phosphorous reacting

and, For 1.40 moles P2O5 in Part B, we had 1.40 moles of Phosphorous reacting

Similarly we can do the calulations for the rest.