A mixture of 156.3 g of P and 161.5 g of O2 reacts completely to form P4O6 and P4O10 . Find the masses of P4O6 and P4O10 that are formed by the reaction.

Answer 1

1st step is to write a chemical equation for the reaction and then balance it.

The unbalanced reaction is:

___ P + ___ O₂ → ___ P₄O₆ + ___ P₄O₁₀

To balance, we apply the Law of Conservation of Matter so the total number of P and O atoms on the left equal the total number of P and O atoms on the right. We use coefficients in the blanks to balance.

8 P + 8 O₂ → 1 P₄O₆ + 1 P₄O₁₀

This means that 8 moles of P reacts with 8 moles of O₂ to form 1 mole of each product.

Now, we need to check to see if either of our reactants are limiting and will run out before reaching completion. Since the reaction proceeds with equimolar amounts, we just need to identify if the starting amount of moles of P and moles of O₂ are the same.

156.3 g P • (1 mole P / 30.97 grams P) = 5.05 moles P

161.5 g O₂ • (1 mole O₂ / 32.00 grams O₂) = 5.05 moles O₂

Since they are the same amount AND the balanced equation requires a 1/8 ratio for both reactants, this will go to completion and we do not need to worry about limiting reactants.

This requires us to perform two sets of gram to mole to mole to gram calculations. The general format for these is to convert grams of substance A to moles of substance A, then use the mole ratio from the balance equation to convert moles of substance A to moles of substance B, then convert moles of substance B back to grams of substance B.

You can start either with the amount of P given OR the amount of O₂ given since it will give the same result.

Try the calculation yourself starting with 161.5 grams of O₂ and see if you get the same answers.

156.3 g P • (1 mole P / 30.97 grams P) • (1 mol P₄O₆ / 8 mole P) • (219.88 grams P₄O₆ / 1 mole P₄O₆ ) = 138.7 grams P₄O₆

156.3 g P • (1 mole P / 30.97 grams P) • (1 mol P₄O₁₀ / 8 mole P) • (283.88 grams P₄O₁₀ / 1 mole P₄O₁₀)= 179.1 grams P₄O₁₀