NAME Exercise 9 SECTIONDATE Stoichiometry Show caleulation setupe and anawers fo
ID: 1041175 • Letter: N
Question
NAME Exercise 9 SECTIONDATE Stoichiometry Show caleulation setupe and anawers for all problems 1. Use the equation given to solve the following problemE (a) How many moles of Na,PO, would be required to react with LO mol of AGNO,T (b) How many moles of NaNO, can be produced from 0.50 mol of Na,PO (c) How many grams of Ag,PO, can be produced from 6.00 g of Na PO, (d) If you have 9.44 g of Na,PO, how many grams of AgNO, will be needed Sor coemplete reaction (e) When 25.0 g of AgNO, are reacted with excess Na,PO.18.7 of Ag,PO, are pro- duced. What is the percentage yield of Ag,PO 123Explanation / Answer
Ans. Part 1: #a. Following stoichiometry of balanced reaction, 1 mol Na3PO4 reacts with 3 mol AgNO3.
So,
Required moles of Na3PO4 = (1/3) x Moles of AgNO3 = (1/3) x 1 mol = 0.333 mol
#b. Following stoichiometry of balanced reaction, 1 mol Na3PO4 forms 3 mol NaNO3.
So,
Moles of NaNO3 produced = 3 x Moles of Na3PO4 = 3 x 0.5 mol = 1.5
#c. Moles of Na3PO4 = Mass / MW =5.00 g / (163.941 g/ mol) = 0.0305 mol
# Following stoichiometry of balanced reaction, 1 mol Na3PO4 forms 1 mol Ag3PO4.
So,
Moles of Ag3PO4 produced = Moles of Na3PO4 = 0.0305 mol
Now,
Mass of Ag3PO4 produced = Moles x MW = 0.0305 mol x (418.576) = 12.77 g
#d. Moles of Na3PO4 = Mass / MW = 9.44 g / (163.941 g/ mol) = 0.0576 mol
Required moles of AgNO3 = 3 x Moles of Na3PO4 =3 x 0.0576 mol = 0.1728 mol
Now,
Required mass of AgNO3 = 0.1728 mol x (169.873 g/ mol) = 29.354 g
#e. Moles of AgNO3 = 25.0 g / (168.873 g/ mol) = 0.1480 mol
Following stoichiometry, theoretical moles Ag3PO4 formed = (1/3) x Moles of AgNO3
= (1/3) x 0.1480 mol = 0.04933 mol
Theoretical yield of Ag3PO4 = 0.04933 mol x (418.576 g/ mol) = 20.65 g
Now,
% yield = (Actual yield / Theoretical yield) x 100
= (18.7 g / 20.65 g) x 100
= 90.56 %
# Part 2: #a. Moles of KMnO4 = 35 g / (158.034 g/ mol) = 0.2215 mol
Following stoichiometry, required moles of HCl = (16 / 2) x Moles of KMnO4
= (16 / 2) x 0.2215 mol
= 3.544 mol
#b. Following stoichiometry, moles of Cl2 formed = (5 / 2) x Moles of KMnO4
= (5 / 2) x 3.0 mol
= 7.5 mol
#c. Moles of MnCl2 = 35.0 g / (125.84345 g/ mol) = 0.2781 mol
# Following stoichiometry, moles of HCl required = (16 / 2) x Moles of MnCl2
= (16/ 2) x 0.2781 mol = 0.5562 mol
Now,
Required mass of HCl = 0.5562 mol x (36.46064 g/ mol) = 20.28 g
#d. Following stoichiometry, moles of H2O formed = (8 / 2) x Moles of KMnO4
= (8/ 2) x 8.0 mol = 32.0 mol
#e. Moles of KMnO4 = 70.0 g / (158.034 g/ mol) = 0.4429 mol
Moles of HCl = 15.0 g / (36.46064 g/ mol) = 0.4114 mol
# Theoretical molar ratio of reactants = KMnO4 : HCl = 2 :16 = 1 : 8
# Experimental molar ratio of reactants = KMnO4 : HCl = 0.4429 : 0.4114 = 1 : 0.93
Comparing the theoretical and experimental molar ratio of reactants, the experimental moles of HCl is less than its theoretical value of 8 mol while keeping that of KMnO4 constant at 1 mol.
So, HCl is the limiting reactant.
# Formation of product follows the stoichiometry of limiting reactant.
Now,
Theoretical moles of Cl2 formed = (5 / 16) x moles of HCl
= (5 / 16) x 0.4114 mol = 0.1285625 mol
And.
Theoretical (maximum) mass of Cl2 formed = 0.1285625 mol x (70.9054 g/ mol)
= 9.12 g
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