CHEM 1211-Chapter 3 Worksheet Limiting Reactants 1. A weighed sample of phosphor
ID: 1085169 • Letter: C
Question
CHEM 1211-Chapter 3 Worksheet Limiting Reactants 1. A weighed sample of phosphorous (Pa) is added to 17.0 g of chlorine (Cla) and is allowed to react completely. The reaction produces a single product, which can be isolated and weighed. The reaction was repeated several times with different amounts of phosphorous but always using 17.0 g of chlorine. The results are summarized in the graph below 25 20 10 Mass of Ps(g a) What is the limiting reactant when 6.0 g of Ps is added? b) How many grams of P4 must be added to reach a stoichiometric ratio of P4 to Ch? c) What is the mole ratio of Ps to Cl in the reaction? (show your work) mali o of P, to Cl? What is the empirical formula of the produet? d) e) Write the balanced chemical equation for the reaction. ft) Give the systematic name of the product of this reaction. g) What is the % yield of this reaction?Explanation / Answer
a) After mass of P4 is 5g the reaction rate remains constant independent of increase in initial mass of P4.This is because the rate then is limited by other reagent Cl2 amount. So Cl2 is limiting reagent when P4=6g.
b)Till initial mass of P4 becomes 5g rate increases with initial mass of P4 increase then becomes constant,after that Cl2 becomes limiting reagent.This means when P4=5g, Cl2 and P4 are both in stoichiometric amounts.Thus answer is 5g.
c)Stoichiometric amounts are 5g P4 and 17g of Cl2. Molar ratio of P and Cl= [(5g/(3.97g/mol)]/[17g/(35.45g/mol)] =0.2892 =1/3.458=2/6.92=2/7(approx.)
d)Empirical formula is P2Cl7.
e). P4 + 7Cl2 ===>2P2Cl7
f) Diphosphorus Heptachloride
g)With 5g of P4 and 17g of Cl2 reacting, 21g of product should be formed theoritically.But actual formed is 20g(see plot for 5g of P4).So %yield = (Actual/Theoritical)*100=(20/21)*100=95.24%
Comment in case of any doubt.