Question

Determination of theoretical yield NaCl: volume HCl used (L) molarity HCl (mol/L) moles HCl available to react Type answer here 1.998 volume NaOH used (L) molarity NaOH (mol/L) 2.016 moles NaOH available to react Type answer here Theoretical yield of NaCl, expressed as moles NaCl: Type answer here

Answer 1

We know that, Molarity = No. of moles of solute / Volume of solution in L

$\therefore$ No. of moles of solute = Molarity $\times$ volume of solution in L

No. of moles of HCl = 1.998 mol / L $\times$ 0.05 L = 0.0999 mol

No. of moles of NaOH = 2.016 mol / L $\times$ 0.05 L = 0.1008 mol

Consider a reaction, HCl + NaOH $\rightarrow$ NaCl + H₂O  

From reaction, 1 mol HCl $\equiv$ 1 mol NaOH $\equiv$ 1 mol NaCl

We have, No. of moles of HCl = 0.0999 mol and No. of moles of NaOH = 0.1008 mol

Hence, Moles of HCl are less than moles of NaOH. Here HCl is limiting reactant and yield of NaCl will depend on amount of HCl.

We have relation, 1 mol HCl $\equiv$ 1 mol NaCl

$\therefore$ 0.0999 mol HCl $\equiv$ 1 $\times$ 0.0999 / 1 mol NaCl

0.0999 mol HCl $\equiv$ 0.0999 mol NaCl

i e Theoretical yield of NaCl = 0.0999 mol

ANSWER :

Volume of HCl used	Molarity of HCl	moles of HCl available to react
0.05 L	1.998	0.0999
Volume of NaOH used	Molarity of NaOH ( mol/L)	moles of NaOH available to react
0.05 L	2.016	0.1008
Theoretical yield of NaCl		0.0999 moles