Please show all work.
- In a particular reaction between copper metal and silver nitrate, 54.0 g AgNO3 produced 25.0 g Ag. What is the percent yield of silver in this reaction?
Cu +2AgNO3 --> Cu(NO3)2 + 2Ag
- Ammonia reacts with oxygen according to the equation below.
a. If 0.23 mol NH_3 reacts with 0.19 mol O_2, which is the limiting reactant?
b. Calculate the theoretical yield (in grams) of nitrogen, N_2.
1. Percent yield is actual divided by theoretical. There is no way that 25g of Ag could have been produced because theoretically, 17.14g of Ag is produced. 25/17.14 is greater than one; it should be less than one to give you a percentage.
2. To find the limiting reactant in this problem, you multiply .23 by 4 and .19 by. The result is .92 mol of NH3 and .57 mol of O2. Therefore, O2 is the limiting. From this amount of O2, you can find how much in theory grams of N2 should be produced. Multiply .19mol O2 by 3mol of O2 and then multiply 2mol of N2 by 28g of N2, divide this number by 3mol of O2. You should get 10.64 grams of N2. Now you can find the theoretical yield