Stoichiometry Pre-Lab Assignment: 1. Describe the procedure for vacuum filtratio
ID: 1087741 • Letter: S
Question
Stoichiometry Pre-Lab Assignment: 1. Describe the procedure for vacuum filtration.(3 pts) Name: Date MUST SHOW ALL WORK TO GET FULL CREDIT. USE SIGNIFICANT DIGITS A student performed the following reaction using 2.50 mL of a 5.00 M aqueous solution of silver nitrate (AgNO,) and 1.80 mL of 3.00 M potassium chromate (K CrOs) and obtained 1.567 g of silver chromate (Ag,CrO). Determine the limiting reactant. Calculate the theoretical yield and % yield of A&C04.; Use correct significant digits and show all work for full credit. 2. a) Determine the number of moles of each reactant. (2pts) b) Determine the theoretical moles that each reactant will produce. Then determine which reactant is the limiting reactant. (2 pts) c) Calculate the theoretical yield and % yield of AgCro. (3pts)Explanation / Answer
2 AgNO3 + K2CrO4 ----------------- Ag2 CrO4 + 2 KNO3
a).AgNO3 = 2.50 ml of 5.00M
number of moles of AgNO3 = 5.00M x 0.00250L= 0.0125 mole
K2CrO4= 1.80 ml of 3.00M
number of moles of K2CrO4 = 3.00M x 0.00180L= 0.0054 moles
b)
according to equation
1 mole of K2CrO4 = 2 moles of AgNO3
0.0054 moles of K2CrO4 = ?
= 2 x 0.0054/1 = 0.0108moles of AgNO3
we need 0.0108 moles of AgNO3.. but we have 0.0125 moles of AgNO3.So it is excess reagent.
Hence K2CrO4 is limiting reagent.
According to equation
1 mole of K2CrO4 = 1 mole of Ag2 CrO4
0.0054 moles of K2CrO4 = ?
= 1x0.0054 /1 = 0.0054 moles of Ag2CrO4
number of moles of Ag2CrO4 = 0.0054 moles
number of moles of Theoritical yield of Ag2CrO4= 0.0054 moles
molar mass of Ag2CrO4 =331.73 gram/mole
mass of 0.0054 moles of Ag2CrO4 = 0.0054 x 331.73= 1.79 grams
Theoritical yiled of Ag2CrO4 = 1.79 grams
Actual yiled = 1.567 grams
Percent yiled = Actual yiled/theoritical yiled x100
= 1.567/1.79 x100 = 87.54%
Percent yield = 87.54%