StoichiometryStoichiometry

Throughout this experiment, our class was trying to solve this question, “Which Balanced
Chemical Equation Best Represents the Thermal Decomposition of Sodium Bicarbonate?” Me and my
partners were given four equations that determined with is the products of Sodium bicarbonate that was
given. But to know, we have to do the experiment by boiling the Sodium bicarbonate to see what the
ending product and which equation it was by using Stoichiometry. Using a balanced chemical equation to
calculate amounts of reactants and products is called stoichiometry. It is a super technical-sounding word
that simply means using ratios from the balanced equation. This allows chemists to use a balanced
chemical equation to represent what happens on the submicroscopic level during a chemical reaction
rather than doing the whole experiment. By doing the experiment and doing the stoichiometry, we were
able to know that the known properties of the products of each reaction suggested that equation #2 was
right.
First, have to do stoichiometry to the four equation to see what our ending products is equal to the
grams of the each balanced equation to our ending product to the experiment part of the lab (look at A1
for the Stoichiometry). Then, we needed to get the equipment to start this experiment. For example,
Bunsen burner, Striker, Ring stand with metal ring, Crucible with lid, and Crucible tongs. After that, we
measured 2.5 grams of Sodium bicarbonate and placed that in the crucible, measured it again with the
crucible. After that, we placed the crucible with the solution under a burner and time it to 5 mins. By
doing this, we are allowing all the water of the solution out of it. When the 5 mins were done, we
measured the final mass of the left over Sodium bicarbonate and saw which balanced equation was the
closers to our final grams and we did the same process again to see that both the times, the product is the
same which it was (Look at A2 for the measurements).
Thought out this experiment, many key points play out in this experiment like The Law of
Conservation of Mass states that mass is conserved during a chemical reaction. The law of definite
proportions states that a compound is always made up of the exact same proportion of elements by mass. John Dalton was able to explain these two fundamental laws of chemistry with his atomic theory, which
states that a chemical reaction is simply the rearrangement of atoms with no atoms being destroyed and
no new atoms being produced during the process. After heating the substance we let it cool down and we
put the substance in about 200ml of water. Based on the properties of the possible substances that
remained, there were several ways the water and the unknown salt could have reacted.The substance
didn’t catch on fire so reaction equation couldn’t be the fourth one because NaH reacts strongly with
water. The water didn’t heat up so it couldn’t be the first or the third equation because both NaOH and
Na2O would cause this. We then conducted a Ph test and the water turned purple meaning it had turned
into a base so the reaction had to be one of the first two equation and since the water didn’t heat up so it
had to be the second equation. We conducted the test twice. Based on the properties of the possible solids
that could be produced from the decomposition of Sodium bicarbonate we know that the substance
produced in our experiment was Sodium carbonate from the second possible equation. Which most of the
whole class got that the second balanced equation was the same gram amount as we got in the lab. But
other people got the third balanced equation because they heated the Sodium bicarbonate in the crucible
for a longer time which gave more water molecules to escape. A1:
Stoichiometry for each of the chemical equations:
2.5g NaHCO3 2.5g NaHCO3 1 mol NaHCO3 1 mol NaOH 39.997g NaOH 84.006g
NaHCO3 1 mol NaHCO3 1 mol NaOH 1 mol NaHCO3 1 mol NaCO3 105.988g NaCO3 84.006g 2 mol NaHCO3 1 mol NaCO3 =1.19g NaOH =1.577g
2NaHCO3 NaHCO3
2.5g NaHCO3 2.5g NaHCO3 1 mol NaHCO3 1 mol Na2O 61.979g Na2O 84.006g
NaHCO3 2 mol NaHCO3 1 mol Na2O 1 mol NaHCO3 1 mol NaH 23.998g NaH 84.006g
NaHCO3 1 mol NaHCO3 1 mol NaH =0.922g Na2O =0.714g NaH A2:
Data collected from trials:
Trial
#1 Trial
#2 Initial Sodium Initial
Bicarbonate
Crucible
Mass
Mass Initial Crucible & Crucible & Sodium
Sodium
Bicarbonate Mass
Bicarbonate Mass After Heat Substance
Left 2.5g 18.66g 1.568g 16.16g 17.728g Initial Sodium Initial
Bicarbonate
Crucible
Mass
Mass Initial Crucible & Crucible & Sodium
Sodium
Bicarbonate Mass
Bicarbonate Mass After Heat Substance
Left 2.5g 18.669g 1.545g 16.16g 17.705g

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