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monitta
3 years ago
7

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.
Chemistry
1 answer:
astra-53 [7]3 years ago
8 0

Given:

Ammonia n h_{3}=6.91 g

Hydrochloric acidh 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 acidh 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

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