Bromine, a liquid at room temperature has a boiling point of 58 degrees celsius and a melting point at -7.2 degree celsius bromi
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The given question is incomplete. The complete question is ;
Ammonia gas can be prepared by the following reaction:
In an experiment, 27.6 g of ammonia gas, , is produced when it was predicted that 50.2 g of
would form.
What is the theoretical yield of ? What is the actual yeild of
? What is the percent yeild of
?
Answer: a) 50.2 g
b) 27.6 g
c) 5.0 %
Explanation:
a) Theoretical yield is defined as the amount of the product that is possible in the reaction.
Theoretical yield of = 50.2 g
b) Experimental or actual yield is defined as the amount of the product that is actually formed in the reaction.
Experimental yield of = 27.6 g
c) Percentage yield is defined as the ratio of experimental yileld to the theoretical yield in terms of percentage.
Hence percentage yield for the reaction is 55.0 %.
<u>Answer:</u> The concentration of KOH solution is 0.103 M.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of KHP = 0.4877 g
Molar mass of KHP = 204.22 g/mol
Putting values in equation 1, we get:
The chemical reaction for the reaction of KHP with KOH follows:
By Stoichiometry of the reaction:
1 mole of KHP reacts with 1 mole of KOH.
So, 0.0024 moles of KHP will react with = of KOH.
- To calculate the molarity of KOH, we use the equation:
We are given:
Moles of KOH = 0.0024 moles
Volume of solution = 23.22 mL = 0.02322L (Conversion factor: 1L = 1000 mL)
Putting values in above equation, we get:
Hence, the molarity of KOH solution is 0.103 M.