what mass of PF3 would be produced if 0.5g of P4 and .5g of F2 react with a 78.1

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what mass of PF3 would be produced if 0.5g of P4 and .5g of F2 react with a 78.1% yield? ANSWER: Step 1: Balanced Equation. P4 + 6 F2-------> 4 PF3 Step 2: Finding Limiting reactant. Molar mass of P4 = 124 g/mol Molar mass of F2 = 38 g/mol Molar mass of PF3 = 88 g/mol Let us take 0.5 g of P4. i). Moles of P4 = Mass of P4/Molar mass of P4 = 0.5 g/124 g/mol = 0.004 moles P4. ii). Convert P4 moles to F2 moles. (moles ratio of P4:F2 is 1:6). 0.004 moles P4 x (6 moles F2/ 1 mole P4) = 0.024 moles F2 iii). Convert F2 moles to mass Mass of F2 = Moles of F2 x Molar mass of F2 = 0.024 moles x 38 g/mol = 0.92 g. 0.5g of P4 need 0.92 g of F2. But we have only 0.5g of F2. So 0.5 g F2 is limiting reactant. Step 3: Finding Theoretical yield. We have to do the similar step what we follow on the limiting reactant calculation. Here we have to use F2 and PF3. i) Convert F2 mass to moles. Moles of F2 = 0.5 g/ 38g/mol = 0.013 moles F2 ii) Convert F2 moles to PF3 moles (Moles ratio of F2:PF3 is 6:4). 0.013 moles F2 x (4 moles PF3/ 6 mole F2) = 0.0087 moles PF3 iii) Convert PF3 moles to mass. = 0.0087 moles PF3x 88 g/mol = 0.763 g PF3. 0.763 g is 100% mass of PF3. Step 4: Actual mass of PF3 is 78.1%. % Yield = (Actual Mass/Theoretical Mass) x 100% 78.1 % = (Actual Mass/0.763 g) x100% (78.1% x0.763)/100% = Actual Mass Actual Mass of PF3 = 0.5956 g. The Actual Mass is 0.5956 g PF3 was produced above reaction.

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