Question 1 Consider the following reaction between sulfur trioxide and water: SO

Question

Question 1

Consider the following reaction between sulfur trioxide and water: SO3(g)+H2O(l) ---> H2SO4(aq) A chemist allows 61.5 g of SO3 and 11.2 g of H2O to react. When the reaction is finished, the chemist collects 53.3g of H2SO4.

Part A: Determine the limiting reactant for the reaction.

Part B: Determine the theoretical yield for the reaction.

Part C: Determine the percent yield for the reaction.

Question 2: For the reaction shown, compute the theoretical yield of the product in grams for each of the following initial amounts of reactants.
2Al(s)+3Cl2(g)?2AlCl3(s)

Part A: 5.5 gAl; 19.8 gCl2

Part B: 0.439 gAl; 2.29 gCl2

Explanation / Answer

As from equation SO3(g)+H2O(l) ---> H2SO4(aq)

one mole of SO3 will react with one mole of water to give one mole of H2SO$

Therefore 80 g of SO3 will react with 18 g of water to give 98 g of H2SO4

So, 61.5 g fo SO3 will need 13.83 g of water

However here only 11.2 g of water is present so it will react with only 49.77 g of SO3 to give 60.977 g of sulphuric acid ideally

So theoretical yield is 60.977 g and limiting reagent is water.

However the yield is 53.3 g

So percentage of yield = 53.3 X 100 / 60.977 = 87.41 %

2.

2Al(s)+3Cl2(g)?2AlCl3(s)

as given in equation,

54 g of Al ( 2 moles) will react with 212.7 g ( 3moles) of chlorine to give 266.68 g of AlCl3

Part A: 5.5 gAl; 19.8 gCl2 , thus one g of chlorine will react with 0.253 g of Al

here 19.8 g of chlorine will react with 5.026 g of chlorine to give 24.824 g of AlCl3

Part B: 0.439 gAl; 2.29 gCl2

Similarly

0.439 g of Al will react with 1.729 g of Cl2 to give 2.1680 g of Alcl3

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