Question

QUESTION 8 Tetraphosphorous hexaoxide is formed by the reaction of phosphorous (PA) with oxygen gas. If the reaction of 54.6 g of P4 with 28.6 g of O2 produce 21.0 g of P406, calculate the % yield for the reaction.

Answer 1

Molar mass of P4 = 123.9 g/mol Molar mass of O2 = 32.0 g/mol Molar mass of P406 = 219.9 g/mol Balanced chemical equation: 1 P4 + 3 02 1 P406 Theoretical moles ratio of P4 : 02 Calculating number of moles: Mass of P4 = 54.6 g Moles = Mass / Molar mass Moles of P4 = ( 54.6 / 123.9 ) Moles of P4 = 0.441 mol = 1 : 3 Mass of O2 = 28.6 Moles = Mass / Molar mass Moles of 02 = ( 28.6 / 32.0 Moles of 02 = 0.894 mol Experimental moles ratio of P4 : ) 02 = = 0.441 : 0.894 1.000 : 2.028

0.894 Limiting reactant is 02 Excess reactant is P4 Mass of products depends upon limiting reactant From balanced equation: 3 moles of 02 produces 1 moles of P406 0.894 moles of 02 produces ? moles of P406 moles of P406 = (1/ 3 )* moles of P406 = 0.298 mol Calculating mass: Mass = Moles x Molar mass mass of P406 = 0.298 × 219.9 mass of P406 = 65.5 g Theoretical Yield of P406 = 65.5 g Calculating Percent Yield: Actual Yield of P406 = 21.0 g Theoretical Yield of P406 = 65.5 g Actual Yield Percent Yield = x 100 Theoretical Yield 21.0 Percent Yield of P406 = - X 100 65.5 Percent Yield of P406 = 32.1 %