Question

3. Consider the following UNBALANCED equation: Na (s) Cl2(g) NaCl (s) + → a. Calculate the limiting reagent when 55.0 g of Na react with 67.2 g of CL, (Show all work, and clearly state which is the limiting reagent.)

Answer 1

Solution:- (3) (a) Left side has two Cl and right side has only one. So, to make them balance we need to multiply NaCl by 2 and then to balance Na, we need to multiply by 2.

2Na(s) + Cl₂(g) -------------> 2NaCl(s)

mass of Na = 55.0 g

mass of Cl₂ = 67.2 g

to figure out which reagent is limiting we calculate the mass of product, limiting reagent is the one that gives less amount of the product.

how we do this is...

grams of reactant --------> moles of reactant -------> moles of product -----> grams of product

grams to moles is done by dividing the grams by molar mass. moles to moles is done by multiplying the moles by molar mass and then finally moles to grams is done by multiplying the moles of molar mass.

55.0 g of Na x (1 mol of Na/23 g of Na) x (2mol of NaCl/2 mol of Na) x (58.5 g of NaCl/1mol of NaCl) = 139.89 g of NaCl

67.2 g of Cl₂ x (1mol of Cl₂/71.0 g of Cl₂) x (2mol of NaCl/1mol of Cl₂) x (58.5 g of NaCl/1 mol of NaCl) = 110.74 g of NaCl

From calculations we get the less amount of the product by Cl₂, so the limiting reagent is Cl₂.

(b) Theoretical yield is 110.34 g of NaCl.

(c) percent yield = (actual yield/theoretical yield) x 100

Let's plug in the values in it.

percent yield = (105/110.34) x 100 = 95.2%