Eighty grams (80.0g) Potassium chlorate are mixed with 59.5g Hydrochloric acid and allowed to react according to the balanced equation (formulate into a balanced chemical equation). The amount of Chlorine produced is 18.7g. Calculate the % yield of Chlorine.
Answers
Answer:
please make me brainlist
Explanation:
hu ki baba hi di hai tt I think you will fin
Eighty grams (80.0g) Potassium chlorate are mixed with 59.5g Hydrochloric acid and allowed to react according to the balanced equation. The amount of Chlorine produced is 18.7g.
We have to find the percentage yield of chlorine.
Reaction with potassium chlorate and hydrochloric acid :
potassium chlorate reacts with hydrochloric acid and forms potassium chloride, chlorine and water. the balanced chemical reaction of it is given as, KClO₃ + 6HCl ⇒ KCl + 3Cl₂ + 3H₂O
here you see, one mole of potassium chlorate reacts with 6 moles of hydrochloric acid and gives 3 moles of chlorine.
now let's find the no of moles of potassium chlorate and hydrochloric acid from given mass.
we know, no of moles = given mass/molar mass
so no of moles of KClO₃ = mass of KClO₃/molar mass of KClO₃
= 80/122.5 = 0.653
no of moles of HCl = mass of HCl/molar mass of HCl
= 59.5/36.5 = 1.63
from reaction, one moles of KClO₃ reacts with 6 moles of HCl.
∴ 0.653 mol of KClO₃ will react with 6 × 0.653 = 3.918 mol of HCl. but here only 1.63 mol of HCl is present hence, HCl is limiting reagent.
so the product will form according to HCl.
we see from chemical reaction, 6 mol of HCl produce 3 mol of chlorine gas.
∴ 1.63 mol of HCl will produce 0.815 mol of chlorine gas.
hence, the mass of 0.815 mol of chlorine gas = no of moles of Cl₂ × molar mass of Cl₂
= 0.815 × 71 = 57.865g
∵ percentage yield = experimental value/theoretical value × 100
here, experimental value of amount of chlorine gas = 18.7 g
while theoretical value of chlorine gas = 57.865 g
∴ percentage yield = 18.7/57.865 × 100 = 32.31 %