Given:
Ammonia [tex]n h_{3}=6.91 g[/tex]
Hydrochloric acid[tex]h c l=4.61 g[/tex]
To find:
The amount excess of reactant left over in ammonia.
Solution:
Ammonia reacts rapidly with hydrochloric acid to form Ammonium Chloride
Equation for the above statement is derived as:
[tex]n h_{3}+h c l=n h_{4} c l[/tex]
One gram per mole of ammonia [tex]n h_{3}=\frac{17 g}{m o l e}[/tex]
Similarly for 6.91g of [tex]n h_{3}=\frac{\frac{6.91}{17 g}}{m o l}[/tex]
[tex]=0.40645 \text { moles of } n h_{3}[/tex]
One gram per mole of hydrochloric acid[tex]h c l=\frac{36.45 g}{\text {mole}}[/tex]
Similarly for 4.61g of [tex]h c l=\frac{\frac{4.61}{36.45 g}}{\text {mole}}[/tex]
[tex]=0.12647 \text { moles of } h c l[/tex]
From the above information we can say that h c l is a limiting reactant.
Limiting reactant is an element that consumes lesser product in a chemical reaction.
Thus the amount of excess reactant is calculated by using the following formula
Amount of excess reactant left over in [tex]n h_{3}=\text {moles in 6.92 g of } n h_{3}-\text { moles in 4.61 } g \text { of } h c l[/tex]
[tex]=0.40645-0.12647[/tex]
[tex]=0.27998 m o l e s \times \frac{17 g}{m o l e}[/tex]
[tex]=4.75966 g[/tex]
Result:
Amount of excess reactant left over ammonia [tex]n h_{3}=4.75966 g[/tex]
