BuyQualityEssays.com – A reaction is performed in a lab whereby two solutions are mixed together. The products are a liquid and a solid precipitate. What procedures would facilitate measurement of actual yield of the solid?

A reaction is performed in a lab whereby two solutions are mixed together. The products are a liquid and a solid precipitate. What procedures would facilitate measurement of actual yield of the solid?

Calculate the maximum potential yield, and apply the final value in the appropriate equation. 

Use filter paper to isolate and dry the precipitate, and then determining mass of the solid. 

Find the cumulative mass of the precipitate and solid, and extrapolate the molecular mass.

All of the above
FEEDBACK
This question focuses on Learning Objective 7: Perform a precipitation reaction and measure the precipitate to calculate percent yield.
Question 8
Why might the actual yield of a precipitation reaction not equal the theoretical yield?

Environmental conditions may cause variations in the water content of the hydrate.

The reaction did not go to completion.

The precipitate was not properly isolated.

All of the above
FEEDBACK
This question focuses on Learning Objective 8: Explain differences between theoretical and actual yield in a controlled experiment.
Extension Questions
Question 1
A student carries out the precipitation reaction shown below, starting with 0.030 moles of calcium nitrate. The final mass of the precipitate is 2.9 g. Answer the questions below to determine the percent yield.
3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)
a. Which product is the precipitate?
b. How many moles of the precipitate would one expect to be produced from 0.030 moles of calcium nitrate?
c. How many grams of solid do you expect to be produced?
d. What is the percent yield?
a. A solid compound is formed in the reaction is the precipitate. The precipitate is Calcium phosphate Ca3(PO4)2  
 
b. 3 moles of Ca(NO3)2 produces 1 moles of Ca3(PO4)2 
 
0.030 mol of Ca(NO3)2 would produce 0.030mol * 1/3 = 0.010 mol of a3(PO4)2 
 
4 Formed precipitate moles are 0.010 g/mol
 
Molecular weight of Ca3(PO4)2 = 310.18 g/mol
 
Theoretical Yield = moles * molecular weight  = 0.010 * 310.18 g/mol = 3.1018 g
 
c. The expectation is to produce a mass of solid of 3.10 g
 
   d. Actual Yield is 2.9 g
 
Precent Yield = (Actual Yield / Theoretical Yield) * 100 
 
Precent Yield = 2.9 g / 3.10 g * 100 = 93.49
 
Precent Yield = 93.49
 
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