1. Show that (Eq. 3) of our synthesis involves an oxidation of the copper by ass
ID: 547313 • Letter: 1
Question
1. Show that (Eq. 3) of our synthesis involves an oxidation of the copper by assigning the appropriate oxidation states, and writing out each half reaction 2. Eq. of our synthesis involves both an oxidation and a reduction of copper. Show this by appropriate assignment of oxidation states, and writing out each half reaction 1. Write out the procedure in your own words,in enough detail that you could use it as your set of directions. 2. State a hypothesis to answer each of the questions we will be addressing during the lab How does solution volume during crystallization affect % yield? How does solution temperature during crystallization affect % yield? a. b. 3. Develop a balanced equation for the overall reaction by adding up equations 3 through 8. (Note the need to multiply equation 8 by 2, and cancel anything that's the same on both sides of the equation after adding) 4. Do the following calculations, showing your work a. If 1g of Cu is used in our synthesis, how many moles of Nitric Acid are required: b. How many moles Nitric Acid are contained in 5 mL of 15M Nitric Acid? c. Why do you think we use excess nitric acid? d. How many moles Sodium Carbonate are contained in 5g? e. How many did we need to neutralize the excess nitric acid? f. Why do you think we use more Na CO, than needed to neutralize the acid?Explanation / Answer
a)
mol of Cu = mass/MW = 1/63.5 = 0.0157 mol of Cu
Cu + 2HNO3 = Cu+2 + H2
mol of HNO3 = 2*0.0157 = 0.03144 mol of HNO3 required
b)
HNO3 for V = 5 mL of 15 M nitric acid
mol = MV = (15)(5*10^-3) = 0.075 mol of HNO3
c
excess nitric acid is used to avoid metal left
d
Na2CO3 in 5 g
mol = mass/MW = 5/105.9884 = 0.0471
e
mol of HNO3 for mol of Na2CO3
mol of NaCO3 = 2 mol of HNO3
mol of acid = 0.0471*2 = 0.0942
f
Na2CO3 is used in order to ensure total neutralization of acid
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