Question

purpose (lo): to determine the percent composition of sodium bicarbonate (nahco₃) in an impure mixture of sodium bicarbonate by performing a decomposition reaction and stoichiometry.
safety concerns: hot plate can cause damage to the skin if used incorrectly. show caution
necessary ppe: goggles, gloves
methodology:
method/lab skill: off gassing and stoichiometry
procedure: must be detailed
raw data: one of the three reactions below is taking place. balance the reactions and use them later as reference to determine which is the correct reaction based on the stoichiometry.
nahco₃(s) → naoh(s) + co₂(g)
nahco₃(s) → na₂o(s) + co₂(g) + h₂o(g)
nahco₃(s) → na₂co₃(s) + co₂(g) + h₂o(g)
create a table like the one below

	pt. i	pt. ii
mass of beaker + solid	30.98	30.98
mass of solid	35.8	35.7
mass of product after 1st heating (10 min)	4.82	4.32
mass of product after 2nd heating (5 min)	2.78	3.80
mass of product after 3rd heating* (5 min)	2.75	3.79
mass of product	2.74	3.79

analysis:
will be broken down into 2 categories: calculations & written
pt. i calculations: plug in your mass of solid for the sodium bicarbonate in the data table to determine the mass of product that should be made for each balanced chemical reaction. this will be 3 step stoichiometry problems (mass to mass)
pt. i written: explain the math you are performing and how you can use it to determine which chemical reaction is actually occurring

Explanation:

Step1: Balance the chemical reactions

1. For $NaHCO_3(s)

ightarrow NaOH(s)+CO_2(g)$:

• The balanced equation is $2NaHCO_3(s)

ightarrow Na_2CO_3(s)+H_2O(g)+CO_2(g)$

1. For $NaHCO_3(s)

ightarrow Na_2O(s)+CO_2(g)+H_2O(g)$:

• The balanced equation is $2NaHCO_3(s)

ightarrow Na_2O(s)+2CO_2(g)+H_2O(g)$

1. For $NaHCO_3(s)

ightarrow Na_2CO_3(s)+CO_2(g)+H_2O(g)$:

• It is already balanced as $2NaHCO_3(s)

ightarrow Na_2CO_3(s)+CO_2(g)+H_2O(g)$

Step2: Calculate the mass - to - mass for the first reaction in Pt. I

1. The molar mass of $NaHCO_3$ is $M_{NaHCO_3}=22.99 + 1.01+12.01 + 3\times16.00=84.01\ g/mol$
2. The molar mass of $Na_2CO_3$ is $M_{Na_2CO_3}=2\times22.99+12.01 + 3\times16.00 = 105.99\ g/mol$
3. Given the mass of $NaHCO_3$ (solid) is $m_{NaHCO_3}=4.82\ g$

• The number of moles of $NaHCO_3$, $n_{NaHCO_3}=\frac{m_{NaHCO_3}}{M_{NaHCO_3}}=\frac{4.82\ g}{84.01\ g/mol}\approx0.0574\ mol$
• From the balanced equation $2NaHCO_3(s)

ightarrow Na_2CO_3(s)+H_2O(g)+CO_2(g)$, the mole ratio of $NaHCO_3$ to $Na_2CO_3$ is $2:1$

• The number of moles of $Na_2CO_3$ produced, $n_{Na_2CO_3}=\frac{1}{2}n_{NaHCO_3}=\frac{1}{2}\times0.0574\ mol = 0.0287\ mol$
• The mass of $Na_2CO_3$ produced, $m_{Na_2CO_3}=n_{Na_2CO_3}\times M_{Na_2CO_3}=0.0287\ mol\times105.99\ g/mol\approx3.04\ g$

Answer:

The balanced equations are:

1. $2NaHCO_3(s)

ightarrow Na_2CO_3(s)+H_2O(g)+CO_2(g)$

1. $2NaHCO_3(s)

ightarrow Na_2O(s)+2CO_2(g)+H_2O(g)$

1. $2NaHCO_3(s)

ightarrow Na_2CO_3(s)+CO_2(g)+H_2O(g)$
The mass of $Na_2CO_3$ produced from $4.82\ g$ of $NaHCO_3$ in the first reaction is approximately $3.04\ g$