purpose (lo): to determine the percent composition of sodium bicarbonate (nahco₃) in an impure mixture of…

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.\nsafety concerns: hot plate can cause damage to the skin if used incorrectly. show caution\nnecessary ppe: goggles, gloves\nmethodology:\nmethod/lab skill: off gassing and stoichiometry\nprocedure: must be detailed\nraw 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.\nnahco₃(s) → naoh(s) + co₂(g)\nnahco₃(s) → na₂o(s) + co₂(g) + h₂o(g)\nnahco₃(s) → na₂co₃(s) + co₂(g) + h₂o(g)\ncreate a table like the one below\n| | pt. i | pt. ii |\n|--|--|--|--|\n| mass of beaker | sodium bicarbonate (4 - 5g sample) | sodium bicarbonate mixture (4 - 5g sample) |\n| mass of beaker + solid | 30.98 | 30.98 |\n| mass of solid | 35.8 | 35.7 |\n| mass of product after 1st heating (10 min) | 4.82 | 4.32 |\n| mass of product after 2nd heating (5 min) | 2.78 | 3.80 |\n| mass of product after 3rd heating* (5 min) | 2.75 | 3.79 |\n| mass of product | 2.74 | 3.79 |\nanalysis:\nwill be broken down into 2 categories: calculations & written\npt. 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)\npt. i written: explain the math you are performing and how you can use it to determine which chemical reaction is actually occurring

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.\nsafety concerns: hot plate can cause damage to the skin if used incorrectly. show caution\nnecessary ppe: goggles, gloves\nmethodology:\nmethod/lab skill: off gassing and stoichiometry\nprocedure: must be detailed\nraw 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.\nnahco₃(s) → naoh(s) + co₂(g)\nnahco₃(s) → na₂o(s) + co₂(g) + h₂o(g)\nnahco₃(s) → na₂co₃(s) + co₂(g) + h₂o(g)\ncreate a table like the one below\n| | pt. i | pt. ii |\n|--|--|--|--|\n| mass of beaker | sodium bicarbonate (4 - 5g sample) | sodium bicarbonate mixture (4 - 5g sample) |\n| mass of beaker + solid | 30.98 | 30.98 |\n| mass of solid | 35.8 | 35.7 |\n| mass of product after 1st heating (10 min) | 4.82 | 4.32 |\n| mass of product after 2nd heating (5 min) | 2.78 | 3.80 |\n| mass of product after 3rd heating* (5 min) | 2.75 | 3.79 |\n| mass of product | 2.74 | 3.79 |\nanalysis:\nwill be broken down into 2 categories: calculations & written\npt. 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)\npt. i written: explain the math you are performing and how you can use it to determine which chemical reaction is actually occurring

Answer

Explanation:

Step1: Balance the chemical reactions

  1. For $NaHCO_3(s)\rightarrow NaOH(s)+CO_2(g)$:
    • The balanced equation is $2NaHCO_3(s)\rightarrow Na_2CO_3(s)+H_2O(g)+CO_2(g)$
  2. For $NaHCO_3(s)\rightarrow Na_2O(s)+CO_2(g)+H_2O(g)$:
    • The balanced equation is $2NaHCO_3(s)\rightarrow Na_2O(s)+2CO_2(g)+H_2O(g)$
  3. For $NaHCO_3(s)\rightarrow Na_2CO_3(s)+CO_2(g)+H_2O(g)$:
    • It is already balanced as $2NaHCO_3(s)\rightarrow Na_2CO_3(s)+CO_2(g)+H_2O(g)$

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 $2NaHCO_3(s)\rightarrow Na_2CO_3(s)+H_2O(g)+CO_2(g)$, 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. $2NaHCO_3(s)\rightarrow Na_2CO_3(s)+H_2O(g)+CO_2(g)$
  2. $2NaHCO_3(s)\rightarrow Na_2O(s)+2CO_2(g)+H_2O(g)$
  3. $2NaHCO_3(s)\rightarrow Na_2CO_3(s)+CO_2(g)+H_2O(g)$ The mass of $Na_2CO_3$ produced from $4.82\ g$ of $NaHCO_3$ in the first reaction is approximately $3.04\ g$