Suppose a 48. L reaction vessel is filled with 1.6 mol of Br2 and 1.6 mol of OCl2. What can you say about the composition of the
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Answer:
Percent yield = 92.5%
Explanation:
The question asks for the percent yield which can be defined as:
Where the actual yield is <em>how much product was obtained</em>, in this case 6.11 g of Ca(OH)₂, and the theoretical yield is <em>how much product could be obtained with the given reactants theoretically</em>, that is if the reaction would work perfectly. So we need to calculate first the theoretical yield.
1. First lets write the chemical equation reaction correctly and check that it is balanced:
CaO + H₂O → Ca(OH)₂
2. Calculate the amount of product Ca(OH)₂ that can be obtained with the given reactants (theoretical yield), which are 5.00g of CaO and excess of water. So the amount of CaO will determined how much Ca(OH)₂ we can obtained.
For this we'll use the molar ratio between CaO and Ca(OH)₂ which we see it is 1:1. For every mol of CaO we'll obtain a mol of Ca(OH)₂. So lets convert the 5.00 g of CaO to moles:
Molar Mass of CaO: 40.078 + 15.999 = 56.077 g/mol
moles of CaO = 5.00 g / 56.077 g/mol = 0.08916 moles
As we said before from the molar ratio moles of Ca(OH)₂ = moles of CaO
So the moles of Ca(OH)₂ that can be obtained are 56.077 g/mol
We need to convert this value to grams:
Molar Mass of Ca(OH)₂ = 40.078 + (15.999 + 1.008)*2 = 74.092 g/mol
Theoretical yield of Ca(OH)₂ = 0.08916 moles x 74.092 g/mol = 6.606 g
3. Calculate the percent yield:
Percent yield = (6.11 g / 6.606g) x 100 = 92.5 %