Question

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	30.98	30.98
mass of beaker + solid	35.8	35.3
mass of solid	4.82	4.32
mass of product after 1st heating (10 min)	2.78	3.80
mass of product after 2nd heating (5 min)	2.75	3.79
mass of product after 3rd heating* (5 min)	2.74	3.79
mass of product	2.74	3.79

analysis:
pt. ii calculations: use stoichiometry and mass percentage equations to determine the mass percentage of nahco₃ present in the mixture based on results from experimentation
pt. ii written: thoroughly explain how your equations and calculations prove your percent composition of the mixture
conclusion: discuss what you can conclude from the experiment. account for a 5% error from the actual values. look your instructor for the values. what can you conclude from your findings? what errors occurred in your lab that could be minimized in the future?

Explanation:

Step1: Balance the chemical reactions

1. \(2NaHCO_3(s)

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

• Sodium: On the left - hand side, we have 2 Na in \(2NaHCO_3\), and on the right - hand side, 2 Na in \(Na_2CO_3\).
• Hydrogen: 2 H in \(2NaHCO_3\) and 2 H in \(H_2O\).
• Carbon: 2 C in \(2NaHCO_3\) and 2 C (1 in \(Na_2CO_3\) and 1 in \(CO_2\)).
• Oxygen: 6 O in \(2NaHCO_3\) and 6 O (3 in \(Na_2CO_3\), 2 in \(CO_2\), and 1 in \(H_2O\)).

1. The reaction \(NaHCO_3(s)

ightarrow NaOH(s)+CO_2(g)\) is not balanced. Balanced, it would be \(2NaHCO_3(s)
ightarrow 2NaOH(s) + 2CO_2(g)\) (but this is not a common reaction for \(NaHCO_3\) decomposition).

1. The reaction \(2NaHCO_3(s)

ightarrow Na_2O(s)+2CO_2(g)+H_2O(g)\) is also not a typical decomposition reaction of \(NaHCO_3\). The most common decomposition is \(2NaHCO_3(s)
ightarrow Na_2CO_3(s)+CO_2(g)+H_2O(g)\).

Step2: Calculate mass of \(NaHCO_3\) in the mixture for Pt. II

• Mass of the sample in Pt. II is \(m_{sample}=35.3 - 30.98=4.32g\).
• Mass of the final product in Pt. II is \(m_{product}=3.79g\).
• Let's assume the reaction is \(2NaHCO_3(s)

ightarrow Na_2CO_3(s)+CO_2(g)+H_2O(g)\).

• The molar mass of \(NaHCO_3\) is \(M_{NaHCO_3}=23 + 1+12 + 3\times16=84g/mol\).
• The molar mass of \(Na_2CO_3\) is \(M_{Na_2CO_3}=2\times23+12 + 3\times16 = 106g/mol\).
• From the balanced equation, 2 moles of \(NaHCO_3\) produce 1 mole of \(Na_2CO_3\).
• Let the mass of \(NaHCO_3\) in the mixture be \(x\) grams. The mass of \(Na_2CO_3\) produced from \(x\) grams of \(NaHCO_3\) is \(m_{Na_2CO_3}=\frac{106}{2\times84}x\).
• We know the mass of the final product (assumed to be \(Na_2CO_3\)) is 3.79g. So, if \(m_{Na_2CO_3}=3.79g\), then \(x=\frac{2\times84\times3.79}{106}\approx6.03g\) (this is wrong as the sample mass is 4.32g. There is an error in the assumption or data. Let's use the mass - loss method).
• Mass loss in Pt. II is \(m_{loss}=4.32 - 3.79 = 0.53g\). This mass loss is due to the formation of \(CO_2\) and \(H_2O\).
• From the balanced equation \(2NaHCO_3(s)

ightarrow Na_2CO_3(s)+CO_2(g)+H_2O(g)\), for every 2 moles of \(NaHCO_3\) decomposed, the mass loss is \(44 + 18=62g\) (molar mass of \(CO_2\) and \(H_2O\)).

• Let the mass of \(NaHCO_3\) be \(m\). Then \(\frac{m}{2\times84}=\frac{0.53}{62}\).
• Solving for \(m\), we get \(m=\frac{2\times84\times0.53}{62}\approx1.45g\).
• Mass percentage of \(NaHCO_3\) in the mixture \(=\frac{1.45}{4.32}\times100\%\approx33.6\%\)

Answer:

The mass percentage of \(NaHCO_3\) in the mixture (Pt. II) is approximately \(33.6\%\)