Tetraphosphorous hexaoxide is formed by the reaction of phosphorous (P4) with oxygen gas. If the reaction of 19.4 g of phosphorus (P4) with 68.3 g of oxygen produces 21.0 g of Tetraphosphorous hexaoxide, calculate the percent yield for the reaction.
a. 34.4
b. 156.5
c. 13.4
d. 86.6
e. 61.0
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To calculate the percent yield for the reaction, we need to compare the actual yield (21.0 g) to the theoretical yield.
First, let's calculate the molar mass of P4:
P = 31 g/mol
P4 = 31 g/mol * 4 = 124 g/mol
Next, let's calculate the molar mass of O2:
O = 16 g/mol
O2 = 16 g/mol * 2 = 32 g/mol
Now, let's calculate the number of moles of P4 and O2:
Number of moles of P4 = 19.4 g / 124 g/mol = 0.156 moles
Number of moles of O2 = 68.3 g / 32 g/mol = 2.135 moles
According to the balanced equation, the stoichiometric ratio between P4 and Tetraphosphorous hexaoxide is 1:1. Therefore, the number of moles of Tetraphosphorous hexaoxide formed will also be 0.156 moles.
Now, let's calculate the theoretical yield of Tetraphosphorous hexaoxide:
The molar mass of Tetraphosphorous hexaoxide is:
P4O6 = (124 g/mol * 4) + (16 g/mol * 6) = 524 g/mol
The theoretical yield of Tetraphosphorous hexaoxide = number of moles * molar mass = 0.156 moles * 524 g/mol = 81.744 g
Finally, let's calculate the percent yield:
Percent yield = (actual yield / theoretical yield) * 100% = (21.0 g / 81.744 g) * 100% = 25.7%
Therefore, the percent yield for the reaction is 25.7%.
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