Question

For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants.
2Li(s)+Cl2(g)→2LiCl(s)

(Please show typed out problem solving. I'm trying to get a grasp of how this works, and I have a hard time reading the hand-written answers,.) Thank you!

Part A

1.0 gLi; 1.0 gCl2

Express your answer as a chemical formula.

Part B

10.5 gLi; 37.2 gCl2

Express your answer as a chemical formula.

Part C

2.85×103gLi; 6.79×103gCl2

Express your answer as a chemical formula.

Answer 1

Let us write the equation once again

2 Li(s) + Cl_2(g) ----------> 2 LiCl_(s)

PART (A)

Weight of Li =1 gram

Weight of Cl₂ = 1 gram

Molas mass of LI=6.94 g/mole

Molar mass of Cl₂ = 35.5(2) = 71 g/ mole

Moles of Li =weight of Li/ Molar mass of Li

                = 1/6.94 = 0.144 moles

Moles of Cl₂ = Weight of Cl₂/ Molar mass of Cl₂

                  = 1/71 = 0.01408 moles

Now per the reaction stiochometery

       2 mole of Licl = 1 moles of Cl₂

   Therefore moles of Cl₂ required for 0.144 moles of Li = (1/2) * 0.144 =0.072 moles

But number of moles of Cl2 available is only 0.01408 moles. Hence it will get completely used up

and Li will reamain in excess

Limiting reactang : Chlorine

Part (B)

Weight of Li =10.5 gram

Weight of Cl₂ = 37.2 gram

Molas mass of LI=6.94 g/mole

Molar mass of Cl₂ = 35.5(2) = 71 g/ mole

Moles of Li =weight of Li/ Molar mass of Li

                = 10.5/6.94 = 1.5129 moles

Moles of Cl₂ = Weight of Cl₂/ Molar mass of Cl₂

                  = 37.2/71 = 0.524 moles

Now per the reaction stiochometery

       2 mole of Licl = 1 moles of Cl₂

Moles of Cl₂ required for 1.5129 moles of Li =(1/2)*1.5129 = 0.75645 moles

But moles of Cl₂ available is only 0.524 moles. Hence it will get completely used up and Li will reamin in exces

Limiting reactang : Chlorine

Part (C)

Weight of Li =2.85*10³ gram

Weight of Cl₂ = 6.79*10³ gram

Molas mass of LI=6.94 g/mole

Molar mass of Cl₂ = 35.5(2) = 71 g/ mole

Moles of Li =weight of Li/ Molar mass of Li

                = 2.85*10³/6.94 = 410.663 moles

Moles of Cl₂ = Weight of Cl₂/ Molar mass of Cl₂

                  = 6.79*10³/71 = 95.6338 moles

Now per the reaction stiochometery

       2 mole of Licl = 1 moles of Cl₂

Moles of Cl₂ required for 410.663 moles of Li =(1/2)*410.663 = 205.3315 moles

But moles of Cl₂ available is only 95.6338 moles. Hence it will get completely used up and Li will reamin in exces

Limiting reactang : Chlorine