Question

The mass of $Al(OH{)}_3$ that was obtained by the reaction of 133.4 g of $AlC{l}_3$ with 40 g of $NaOH$ according to following equation is 13 g.
$AlC{l}_3+NaOH\to Al(OH{)}_3+NaCl$ (not balanced)
Calculate the percentage yield.

(Molar mass of aluminium chloride is 133.34 g/mol, molar mass of $NaOH$ is 40 g/mol and molar mass of aluminium hydroxde is 78 g/ mol respectively)

• A. 50
• B. 100
• C. 75
• D. 65

Answer 1

The correct option is A 50

The balanced equation is:
$AlC{l}_3+3NaOH\to Al(OH{)}_3+3NaCl$

Moles of aluminium chloride = $\frac{133.34}{133.34}=1$
Moles of $NaOH$ = $\frac{40}{40}=1$

According to the balanced chemical equation, 1 mole of aluminium chloride requires 3 moles of $NaOH$ . But the number of moles of $NaOH$ available is 1. Hence, $NaOH$ is the limiting reagent.
Hence, the moles of aluminium hydroxide formed = 0.33

Mass of aluminium hydroxide formed = $0.33\times 78$ = 26 g

Theoretical yield = 26 g
Experimental yield = 13 g

Percentage yield = $\frac{\text{experimental yield}}{\text{theoretical yield}}\times 100=\frac{13}{26}\times 100=50\%$