Determine the theoretical yield of p2o5, when 3.07 g of p reacts with 6.09 g of oxygen in the following chemical equation
4 p + 5 o2 > 2 p2o5

14.2 grams

Explanation:-

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number of particles.

To calculate the number of moles, we use the equation:

given mass of phosphorous (P) = 3.07 g

Molar mass of phosphorous (P) =  31 g/mol

Putting in the values we get:

given mass of oxygen = 6.09 g

Molar mass of oxygen =  32 g/mol

Putting in the values we get:

According to stoichiometry:

4 moles of phosphorous combine with 5 moles of oxygen

Thus 0.1 moles of phosphorous combine with = moles of oxygen

Thus phosphorous acts as limiting reagent as it limits the formation of product and oxygen is the excess reagent.

4 moles of phosphorous gives=  2 moles of

Thus 0.1 moles of phosphorous gives = moles of

mass of

Thus the theoretical yield of is 14.2 grams.

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa

i think that the answer is c