Answer to Question #55699 in General Chemistry for sue

Question #55699
Now consider a situation in which 26.0 g of P4 is added to 59.0 g of Cl2, and a chemical reaction occurs. To identify the limiting reactant, you will need to perform two separate calculations: 1.Calculate the number of moles of PCl5 that can be produced from 26.0 g of P4 (and excess Cl2). 2.Calculate the number of moles of PCl5 that can be produced from 59.0 g of Cl2 (and excess P4). Then, compare the two values. The reactant that produces the smaller amount of product is the limiting reactant.**answer to #1 is 0.839 moles...answer to #2 is 0.333 moles Info needed:What mass of PCl5 will be produced from the given masses of both reactants? Express your answer to three significant figures and include the appropriate units.
Expert's answer
P4 + 10Cl2 = 4PCl5
n(PCl5) = 4*n(P4) = 4*m(P4)/M(P4) = 4*26/124 = 0.839 mol
m(PCl5) = M(PCl5)*n(PCl5) = M(PCl5)*4*n(P4) = M(PCl5)*4*m(P4)/M(P4) = 208.5*4*26/124 = 174.9 g
n(PCl5) = 0.4*n(Cl2) = 0.4*m(Cl2)/M(Cl2) = 0.4*59/71 = 0.332 mol
m(PCl5) = M(PCl5)*n(PCl5) = M(PCl5)*0.4*n(Cl2) = M(PCl5)*0.4*m(P4)/M(P4) = 208.5*0.4*59/71 = 69.3 g

Answers :
1. m(PCl5) = 174.9 (g)
2. m(PCl5) = 69.3 (g)

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!


No comments. Be first!

Leave a comment

Ask Your question

New on Blog