Which of the listed factors can cause the percent yield of a reaction to be less than 100%?
a. There may be competing side reactions.
b. Reactions do not always go to completion.
c. Product can be lost in transferring between containers.
d. All of the above factors can cause the percent yield to be less than 100%.
Calculate the number of moles of water produced when 6 mol of Cu(NO3)2 are produced according to the following balanced chemical equation:
3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)
For #1 I'm inclined to chose d, all of them, except that c is a little confusing. Certainly c is true; however, if you're talking about "a" single reaction then "a" single reaction doesn't require transferring between containers so I'm not sure of the context of the c answer.
For #2. From your previous questions I'm just not sure you understand this moles business and how to calculate moles of one material when given moles of another material. Any one number of moles of one substance in a reaction can be converted to any other substance in the reaction. It works this way. You need two things; i.e., you need the balanced equation and you need a starting material. Here is the way it works. You take the starting material and multiply it by a factor. Where do you get the factor. The factor comes from the coefficients in the balanced equation. So in this question, the balanced equation is
3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)
The starting material in the question is 6 mols Cu(NO3)2. You want to convert that to ? mols H2O. So 6 mols Cu(NO3)2 x (factor) = mols H2O.
The coefficients in the balanced equation are 3 mols for Cu(NO)3 and 4 moles H2O. So the factor CAN be (3 mols Cu(NO3)2/4 mols H2O) OR it can be (4 mols H2O/3 mols Cu(NO3)2). So the only real thinking that must be done in one of these problems is which is the correct factor to use. Let me write both of them down for you. The first way is
6 mols Cu(NO3)2 x (factor) = mols H2O
6 mols Cu(NO3)2 x [3 mols Cu(NO3)2/4 mols H2O)] = ? OR
6 mols Cu(NO3)2 x [4 mols H2O/3 mols Cu(NO3)2] = ?
The answer to which to use is in the units. The first one gives you mols Cu(NO3)2 x mols Cu(NO3)2/mols H2O which is [Cu(NO3)2]^2/H2O. Is that what you want? The second one gives you mols H2O for the units because mols Cu(NO3)2 in the numerator of the first part cancels with mols Cu(NO3)2 in the factor part. This is a no-brainer. You're trying to convert from mols Cu(NO3)2 to mols H2O. Mols H2O is what you want. The second factor above give you mols of H2O. So the correct one to use is obvious. You can make a rule from this if you don't want to go through this exercise every time. Starting material x (ending material/starting material). It always works this way. So 6 mols Cu(NO3)2 x (4 mols H2O/3 mols Cu(NO3)2) = 6 x 4/3 = 8 mols H2O.