NaHCO 3 (aq) + HCl (aq) ------------> NaCl(aq) + H 2 O(l) + CO 2 (g) This experi
ID: 521240 • Letter: N
Question
NaHCO3(aq) + HCl (aq) ------------> NaCl(aq) + H2O(l) + CO2(g)
This experiment was done using Stoichemetry and Gases. An HCl reaction was used with the unknown. The test tube was filled with 4 M HCl within about 2 cm of the top.
A. What is the limiting reagent for this reaction? How do you know?
B. A student accidently added the salt solution to the unknown solution (Flask A) before the reaction, instead of distilled water. Explain how this will affect the student’s results.
C. A student forgot to ensure that the drain line (into Beaker C) is full of the salt solution before proceeding with the reaction. Explain how this will affect the student’s results.
D. This experiment in this lab is performed again, but with sodium carbonate (Na2CO3) instead of sodium bicarbonate. What is the mass % of the unknown if 10.025 grams of the solution generates 175 mL of CO2 when the lab pressure and temperature are 763.5 mmHg and 21.5°C? The balanced equation is Na2CO3(aq) + 2 HCl(aq) à 2 NaCl(aq) + H2O(l) + CO2(g)
Lab Temperature (K) Lab Atmospheric Pressure (mmHg) Mass of Flask (g) Mass of Flask Unknown (g) Mass of Unknown Solution (g) Volume of Salt Solution Displaced (mL) Moles of CO2 Produced (mol) Moles of NaHCO3 in Unknown Solution (mol) Mass of NaHCO3 in Unknown Solution (g) Mass of NaHCO3 for Unknown Solution Average Mass Trial 1 296 758.5 160.772 169.582 8.81 218.5 0.0089732 0.0089732 0.7538 8.56% 7.1% Trial 2 296 758.5 161.135 168.854 7.719 144 0.0059137 0.0059137 0.4968 5.64%Explanation / Answer
A) Limiting reagent will be sodium bicarbonate becouse no. of moles of bicarbonate less than HCl.
B) If we added salt solution then it change the rate of conversion of product due to le chatlier principle ( conc of product salt will increase reaction move backword direction.)
C) In this case reaction again move backword direction due to le chatlier principle due this product formation goes slow.
D) % of sodium carbonate will be 7.7 %,
n= PV/RT= .175*763.5/0.082*760*294.5 = no. of moles of CO2 = moles of sodium carbonate
g= 106*n
% of Na2CO3 = (g/10.025)*100
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