Question

A student reacts 20.0 grams of sodium hydroxide with excess hydrochloric acid, and obtains a

yield of 25.0 grams of sodium chloride. What is the percent yield of sodium chloride?
NaOH + HCI --> NaCl + H20

Answer 1

Percent yield of NaCl = 87.75 %

Step-by-step explanation:

Data given

mass of sodium = 20.0 g

actual yield of Sodium chloride = 25 g

percent yield of sodium chloride = ?

Reaction Given:

NaOH + HCI --> NaCl + H₂O

Solution:

First we have to find theoretical yield.

So, Look at the reaction

NaOH + HCI ---------> NaCl + H₂O

1 mol 1 mol

As 1 mole of NaOH give 1 mole of NaCl

Convert moles to mass

molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

molar mass of NaCl = 23 +35.5 = 58.5 g/mol

Now

NaOH + HCI ---------> NaCl + H₂O

1 mol (40 g/mol) 1 mol (58.5 g/mol)

40 g 58.5 g

So 40 g of NaOH gives 58.5 g of NaCl so how many grams of Sodium chloride will be produced by 20 g of NaOH.

Apply Unity Formula

40 g of NaOH ≅ 58.5 g of NaCl

20 g of NaOH ≅ X g of NaCl

Do cross multiply

g of NaCl = 58.5 g x 20 g / 40 g

g of NaCl = 29.25 g

So the Theoretical yield of NaCl = 29.25 g

Now Find the percent yield of NaCl

Formula Used

percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

percent yield = 25 / 29.2 x 100 %

percent yield = 87.75 %

percent yield of NaCl = 87.75 %