Tetraphosphorus hexaoxide is formed by the reaction of phosphorus with oxygen gas. If a mixture of 75.3 g of phosphorus and 38.7 g of oxygen produce 43.3 g of P4O6, what is the percent yield for the reaction

percentage yield = 48. 82%

Explanation:

Firstly, the chemical equation of the reaction is written and balanced.

P4(g) + O2(g) → P4O6

balanced equation.

P4(s) + 3 O2(g) → P4O6(s)

We have to know the limiting reactant of the reaction  by converting the reactants to mole

molar mass of phosphorus = 4 × 31 = 124 g

moles of phosphorus = mass/molar mass = 75.3/124 = 0.60725806451  mol-rxn

molar mass of oxygen = 32 g

moles of  oxygen = 38.7/ 32 = 1.209375 /  3 mol = 0.403125  mol-rxn

Therefore the limiting reactant is oxygen and it will determine the amount of product

P4(s) + 3 O2(g) → P4O6(s)

mass of 3 moles of oxygen = 3 × 32 = 96 g

molar mass of P4O6 = 124 + 96  = 220  g

if  96 g of oxygen produces 220 g of P4O6

38.7 g of oxygen will produce ? grams of  P4O6

theoretical yield of P4O6 = (38.7 × 220)/96

theoretical yield of P4O6 = 8514 /96

theoretical yield of P4O6 = 88.6875

theoretical yield of P4O6 = 88.69

percentage yield = actual yield/theoretical yield × 100

percentage yield = 43.3/88.69 × 100

percentage yield = 4330/88.69

percentage yield = 48.8217386402

percentage yield = 48. 82%

The percent yield of the reaction is 48.9 %

Explanation:

Step 1: data given

Mass of phosphorus = 75.3 grams

Molar mass of phosphorus = 123.90 g/mol

Mass of oxygen = 38.7 grams

Molar mass oxygen = 32.0 g/mol

Mass of P4O6 produced = 43.3 grams

Step 2: The balanced equation

P4(s) + 3O2(g) → P4O6(s)

Step 3: Calculate moles P4

Moles P4 = mass P4 / molar mass P4

Moles P4 = 75.3 grams / 123.90 g/mol

Moles P4 = 0.608 moles

Step 4: calculate moles O2

Moles O2 = 38.7 grams / 32.0 g/mol

Moles O2 = 1.21 moles

Step 5: Calculate the limiting reactant

For 1 mol P4 we need 3 moles O2 to produce 1 mol P4O6

O2 is the limiting reactant. There will react 1.21 moles. P4 is in excess. There will react 1.21 / 3 = 0.403 moles. There will remain 0.608 - 0.403 = 0.205 moles P4

Step 6: Calculate moles P4O6

For 1 mol P4 we need 3 moles O2 to produce 1 mol P4O6

For 1.21 moles O2 we'll have 1.21 / 3 = 0.403 moles P4O6

Step 7: Calculate mass P4O6

Mass P4O6 = moles P4O6 * molar mass P4O6

Mass P4O6 = 0.403 moles * 219.88 g/mol

Mass P4O6 = 88.6 grams

Step 8: Calculate percent yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (43.3 grams / 88.6 grams ) *100 %

% yield =  48.9 %

The percent yield of the reaction is 48.9 %

l dont know l am s sorry

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