Question

tetraphosphorous hexaoxide is formed by the reaction of phosphorous (p4) with oxygen gas. if the reaction of 75.3 g of p4 with 38.7 g of o2 produce 43.3 g of p4o6, calculate the % yield for the reaction.

Answer 1

the Percentage yield for the reaction = 48.8%

What is Percentage yield ?

The % ratio of the theoretical yield to the actual yield is known as the percent yield. It is calculated as the theoretical yield multiplied by 100% divided by the experimental yield. The percent yield is 100% if the theoretical and actual yields are equal. Because the real yield is frequently lower than the theoretical value, percent yield is typically lower than 100%. This may be due to incomplete or conflicting reactions or sample loss during recovery. If the percent yield is more than 100%, more sample than expected was retrieved from the reaction.

4 P + 3 O2 = P4O6

moles P = 75.3 g / 30.9738 g/mol= 2.43

moles O2 required = 2.43 x 3 / 4=1.82

actual moles O2 = 38.7 g /32 g/mol=1.21 so O2 is the limiting reactant

theoretical moles P4O6 = 1.21 / 3=0.403

theoretical mass P4O6 = 0.403 mol x 219.895 g/mol=88.6 g

% yield = 43.3 x 100/ 88.6 = 48.8 %

the Percentage yield for the reaction = 48.8%

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