Respuesta :
The given question is incomplete. The complete question is:
Aqueous hydrochloric acid (HCl) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H2O). If 1.60 g of sodium chloride is produced from the reaction of 1.8 g of hydrochloric acid and 1.4 g of sodium hydroxide, calculate the percent yield of sodium chloride. Be sure your answer has the correct number of significant digits in it.
Answer: Thus the percent yield of sodium chloride is 78.0%
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} HCl=\frac{1.8g}{36.5g/mol}=0.049moles[/tex]
[tex]\text{Moles of} NaOH=\frac{1.4g}{40g/mol}=0.035moles[/tex]
[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex]
According to stoichiometry :
1 mole of [tex]NaOH[/tex] require = 1 mole of [tex]HCl[/tex]
Thus 0.035 moles of [tex]NaOH[/tex] will require=[tex]\frac{1}{1}\times 0.035=0.035moles[/tex] of [tex]HCl[/tex]
Thus [tex]NaOH[/tex] is the limiting reagent as it limits the formation of product and [tex]HCl[/tex] is the excess reagent.
As 1 mole of [tex]NaOH[/tex] give = 1 mole of [tex]NaCl[/tex]
Thus 0.035 moles of [tex]NaOH[/tex] give =[tex]\frac{1}{1}\times 0.035=0.035moles[/tex] of [tex]NaCl[/tex]
Mass of [tex]NaCl=moles\times {\text {Molar mass}}=0.035moles\times 58.5g/mol=2.05g[/tex]
[tex]{\text {percentage yield}}=\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100\%[/tex]
[tex]{\text {percentage yield}}=\frac{1.60g}{2.05g}\times 100\%=78.0\%[/tex]
Thus the percent yield of sodium chloride is 78.0%
