4HCl(g) +O2(g) >>>> 2H2O(l) + 2Cl2(g) When 63.1g of HCl are allowed to react wit

Question

4HCl(g) +O2(g) >>>> 2H2O(l) + 2Cl2(g) When 63.1g of HCl are allowed to react with 17.2g of O2 41.5g of Cl2 are collected A.) determine the theoretical yield of Cl2 for the reaction B.) determine the precent yield for the reaction 4HCl(g) +O2(g) >>>> 2H2O(l) + 2Cl2(g) When 63.1g of HCl are allowed to react with 17.2g of O2 41.5g of Cl2 are collected A.) determine the theoretical yield of Cl2 for the reaction B.) determine the precent yield for the reaction When 63.1g of HCl are allowed to react with 17.2g of O2 41.5g of Cl2 are collected A.) determine the theoretical yield of Cl2 for the reaction B.) determine the precent yield for the reaction

Explanation / Answer

4HCl(g) +O2(g) >>>> 2H2O(l) + 2Cl2(g)

the first thing to do is to calculate the number of moles available for HCl and Oxygen

moles of hcl = mass / molar mass = 63.1 / 36.5 = 1.72 moles

moles of oxygen = 17.2 / 32 = 0.5375 moles

let´s analyze, 1 mole of O2 needs 4 moles of HCl so 0.5375 need 0.5375 * 4 = 2.15 moles , we see that we have lees ammount than the required one so the HCl is the limiting reactant and we will have an excess of oxygen

let´s perform calculation 4 moles of HCl produces 2 moles of Cl2 so we can say that 2 moles of HCl produces 1 mole of Cl2

1.72 moles of HCl will produce 0.5 * 1.72 = 0.86 moles of Cl2

molar mass of Cl2 = 71

mass of Cl2 = moles * molar mass = 0.86 * 71 = 61.06 grams, this is the theoretical yield, the ammount of mass you should be able to get if everything goes right

you only got 41.5 out f 61.06 possible grams so

41.5 / 61.06 = 0.6796 * 100 = 67.96 %, this is the percent yield, the fraction you got from the 100% available.

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.