Ch 8- Google Chrome 78787593 Secure https://session Chem SP20 17TuAMLab Exercise

Question

Ch 8- Google Chrome 78787593 Secure https://session Chem SP20 17TuAMLab Exercise 8 64 Part A Determine the limiting reactant for the reaction consider the reaction between HCI and or Express your answer as a chemical formula. When 631 gofHClae alowed to eact with 17.2 got O2. 49 4g of Cl are collected Submit My Answers Give Up Part B Determine the theoretical yield of Cla for the reaction Submit My Answers Glve Up Part C Determine the percent yield for the reaction percent yield Submit My Answers Ghe Up

Explanation / Answer

number of moles of HCl = 63.1g / 36.46 g.mol^-1 = 1.73 mole

number of moles of O2 = 17.2g / 32.0 g.mol^-1 = 0.538 mole

from the balanced equation we can say that

4 mole of HCl requires 1 mole of O2 so

1.73 mole of HCl will require

= 1.73 mole of HCl *(1 mole of O2/4 mole of HCl)

= 0.433 mole of O2

but we have 0.538 mole of O2 so O2 is excess reactant

so HCl is limiting reactant

4 mole of HCl produces 2 mole of Cl2 so

1.73 mole of HCl will produce

= 1.73 mole of HCl*(2 mole of Cl2/4 mole of HCl)

= 0.865 mole of Cl2

1 mole of Cl2 = 70.9060 g so

0.865 mole of Cl2 = 61.3g

Therefore, theoretical yield of HCl is 61.3g

percent yield = (actual yield / theoretical yield)*100

percent yield = (49.4 / 61.3)*100 = 80.6 %

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