complete the information please Mass of beaker and copper (II) sulfate pentahydr
ID: 533472 • Letter: C
Question
complete the information please
Mass of beaker and copper (II) sulfate pentahydrate (CuSO_4, 5 H_2O): Mass of beaker without copper (II) sulfate pentahydrate (CuSO_4. 5 H_2O): Mass of copper sulfate pentahydrate (CuSO_4. 5 H_2O): Moles of CuSO_4, 5 H_2O (molar mass of CuSO_4): 5 H_2O = 249.70 g/mol): Ammonia (ammonium hydroxide): Total milliliters of ammonia added: Molarity of ammonia used: Moles of ammonia used: Overall balanced chemical equation: You must create the overall equation here using the equations given in the background section the overall equation must have two reactants used (the solid reactant and the liquid) and the solid final product that you massed out: Using the equation above, determine which reagent was the limiting reagent for the synthesis: Limiting reagent: Theoretical yield in moles of product: Mass of watchglass, filter paper, and product (dried): Mass of watch glass and filter paper: Mass of product (dried): Using the theoretical moles of product (determine above), the balanced chemical equation, and the product (molar mass = 245.75 g/mol), calculate the theoretical mass of product produced: Calculate the percent yield:Explanation / Answer
Ans:- Mass of copper sulphate pentahydrate CuSO4.5H2O = 1.585g
Molar mass of copper sulphate pentahydrate =249.70g
Moles of CuSO4.5H2O= Given mass /Molar mass
= 1.585/249.70
= 0.0063M
Here Volume of Ammonia added = 4mm=0.004ml
Molarity=15M
Number of moles of Ammonia = 15x0.004/1000
= 0.00006
The overall equation
CuSO4.5H2O +4NH3 -> [Cu(NH3)4]SO4 + 5H2O
Since NH3 has less number of moles so it is the limiting reagent.
Now theoritical yield = 0.00006 x4
= 0.24g
percent yield = mass of product /theoritical yield x100
= 1.89/0.24 x100
= 78.5%
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