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.

Question

asked Mar 24, 2020 37.5k views

1 Answer

Vadim Shvetsov · Answer 1 · 2020-03-28T11:13:00+0000

Given:

Ammonia

Hydrochloric acid

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:

One gram per mole of ammonia
n h_(3)=(17 g)/(m o l e)

Similarly for 6.91g of
n h_(3)=((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
$h c l=\frac{(4.61)/(36.45 g)}{\text {mole}}$

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

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

Result:

Amount of excess reactant left over ammonia
n h_(3)=4.75966 g