what is the mass of PCL5 produced when 3.30 g pf P4 reacts with 7.89 g of CL2 ga

Question

what is the mass of PCL5 produced when 3.30 g pf P4 reacts with 7.89 g of CL2 gas ? A) Use the standard reaction enthalpies given below to determine AH'rn for the fo 4 points) Pale) + 10 Cl2(g) 4PCIS(s) Given PCIs(s) PCI3(g) +Cl2(g) P4(g) + 6 Cl2(g) 4 PCI3(g) Hn +157 k 30 g f Pa rats wth .89 f a B) What is the mass of PCls produced when 3.30 g of P4 reacts with 7.89g of Ch gas? Moles P4: Moles Cl Moles PCIs: Mass PCIs C) If the actual amount of PCIs produced is 8.05g, what is the percent yield? (2 poin

Explanation / Answer

The reaction is = P4 (g)+ 10 Cl2 (g) -------> 4 PCl5 (s)

so 1 mol of P4 reacts with 10 mol of Cl2 to give 4 mol of PCl5

123.89 gm of P4 reacts with 10 * 70.90 gm of Cl2 ----------> 4 * 208.24 gm of PCl5

So ideally 3.30 gm of P4 will react with 18.151 gm og Cl2

So the limiting reagent is the Cl2

so 709 gm of Cl2 produce 832.96 gm of PCl5

7.89 gm of Cl2 will produce = 832.96 / 709 * 7.89 = 9.269 gm of PCl4

Moles of P4 = 3.30 / 123.89 = 0.0266 moles

Moles of Cl2 = 7.89 / 709 = 0.0111 moles

Moles of PCl5 = 9.269 / 832.96 = 0.0111 moles

Mass of PCl5 = 9.269 gm

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