Question

Ammonia rapidly reacts with hydrogen chloride, making ammonium chloride. Calculate the number of grams of excess reactant when 6.91 g of NH3 reacts with 4.61 g of HCl.

Answer 1

Given:

Ammonia $n h_{3}=6.91 g$

Hydrochloric acid$h c l=4.61 g$

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:

$n h_{3}+h c l=n h_{4} c l$

One gram per mole of ammonia $n h_{3}=\frac{17 g}{m o l e}$

Similarly for 6.91g of $n h_{3}=\frac{\frac{6.91}{17 g}}{m o l}$

                                            $=0.40645 \text { moles of } n h_{3}$

One gram per mole of hydrochloric acid$h c l=\frac{36.45 g}{\text {mole}}$

Similarly for 4.61g of $h c l=\frac{\frac{4.61}{36.45 g}}{\text {mole}}$

                                           $=0.12647 \text { moles of } h c l$

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 $n h_{3}=\text {moles in 6.92 g of } n h_{3}-\text { moles in 4.61 } g \text { of } h c l$

                                            $=0.40645-0.12647$

                                            $=0.27998 m o l e s \times \frac{17 g}{m o l e}$

                                             $=4.75966 g$

Result:

Amount of excess reactant left over ammonia $n h_{3}=4.75966 g$