Theoretcal Yield Example - Armstrong State University Yield Example.pdf · Theoretical Yield...
1
Theoretical Yield Example If 4.50 g of HCl are reacted with 15.00 g of CaCO 3 , according to the following balanced chemical equation, calculate the theoretical yield of CO 2 . 2 2 2 3 CO O H CaCl CaCO HCl 2 + + → + 1. Determine the number of moles of one of the products (CO 2 in this example) produced if all of each reactant is used up. 2 2 CO mol 0616 . 0 HCl mol 2 CO mol 1 x HCl g 5 . 36 HCl mol 1 x HCl g 50 . 4 = 2 3 2 3 3 3 CO mol 1499 . 0 CaCO mol 1 CO mol 1 x CaCO g 1 . 100 CaCO mol 1 x CaCO g 00 . 15 = 2. Use the smallest number of moles of the product (CO 2 ) from step 1 to calculate the theoretical yield of product (CO 2 ). 2 2 2 2 CO g 71 . 2 CO mol 1 CO g 0 . 44 x CO mol 0616 . 0 = Note: Since the reactant, HCl, produces the least amount of product, it is the limiting reactant and the other reactant, CaCO 3 , is in excess. Percent Yield Example If 2.50 g of CO 2 are isolated, after carrying out the above reaction, calculate the percent yield of CO 2 . yield % 3 . 92 % 100 x l theoretica CO g 71 . 2 isolated CO g 50 . 2 2 2 = Notes: If you are given a volume for a reactant, you must determine whether you are working with a pure liquid or a solution. You are probably working with a pure liquid if you are given the starting amount of a reagent in volume without a concentration value. In this case, you need to look up the density; volume substance (mL) x density substance (g/mL) = g substance. Then convert to moles via the MM of the substance. You are definitely working with a solution if you are given the starting amount of a reagent in volume and also a concentration value. a. If the concentration value is molarity (M), volume (L) x M (mol/L) = moles. b. Other concentration definitions must also be used appropriately.
Transcript of Theoretcal Yield Example - Armstrong State University Yield Example.pdf · Theoretical Yield...
Theoretical Yield Example If 4.50 g of HCl are reacted with 15.00 g of CaCO3, according to the following balanced chemical equation, calculate the theoretical yield of CO2.
2223 COOHCaClCaCOHCl2 ++→+ 1. Determine the number of moles of one of the products (CO2 in this example) produced if all of each reactant is used up.
22 COmol0616.0
HClmol2COmol1
xHClg5.36HClmol1
xHClg50.4 =
23
2
3
33 COmol1499.0
CaCOmol1COmol1
xCaCOg1.100
CaCOmol1xCaCOg00.15 =
2. Use the smallest number of moles of the product (CO2) from step 1 to calculate the theoretical yield of product (CO2).
22
22 COg71.2
COmol1COg0.44
xCOmol0616.0 =
Note: Since the reactant, HCl, produces the least amount of product, it is the limiting reactant and the other reactant, CaCO3, is in excess.
Percent Yield Example If 2.50 g of CO2 are isolated, after carrying out the above reaction, calculate the percent yield of CO2.
yield%3.92%100xltheoreticaCOg71.2
isolatedCOg50.2
2
2 =
Notes: If you are given a volume for a reactant, you must determine whether you are working with a pure liquid or a solution. You are probably working with a pure liquid if you are given the starting amount of a reagent in volume without a concentration value. In this case, you need to look up the density; volume substance (mL) x density substance (g/mL) = g substance. Then convert to moles via the MM of the substance. You are definitely working with a solution if you are given the starting amount of a reagent in volume and also a concentration value. a. If the concentration value is molarity (M), volume (L) x M (mol/L) = moles. b. Other concentration definitions must also be used appropriately.
