P4(s)+10Cl2(g)→4PCl5(g)
Now consider a situation in which 29.0 g of P4 is added to 55.0 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 29.0 g of P4 (and excess Cl2).
2. Calculate the number of moles of PCl5 that can be produced from 55.0 g of Cl2 (and excess P4).
3. Then, compare the two values. The reactant that produces the smaller amount of product is the limiting reactant.
1)
Molar mass of P4 = 123.88 g/mol
mass of P4 = 29 g
mol of P4 = (mass)/(molar mass)
= 29/1.239*10^2
= 0.2341 mol
Balanced chemical equation is:
P4(s)+10Cl2(g)→4PCl5(g)
According to balanced equation
mol of PCl5 formed = (4/1)* moles of P4
= (4/1)*0.2341
= 0.9364 mol
Answer: 0.936 mol
2)
Molar mass of Cl2 = 70.9 g/mol
mass of Cl2 = 55 g
mol of Cl2 = (mass)/(molar mass)
= 55/70.9
= 0.7757 mol
Balanced chemical equation is:
P4(s)+10Cl2(g)→4PCl5(g)
According to balanced equation
mol of PCl5 formed = (4/10)* moles of Cl2
= (4/10)*0.7757
= 0.3103 mol
Answer: 0.310 mol
3)
Cl2 produces smaller mol of PCl5
So, Cl2 is limiting reagent
Answer: Cl2
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