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ptownbro
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We have a homework problem that we are not sure is right. Need some help.
The class ran an experiment where they were given 1.99 grams of NaHCO3 and 10.13 grams of HCl to create NaCl, H2O, and CO2. Before starting the experiment, what they were given weighed 12.12 grams in a beaker (which adds up to the 1.99g + 10.13g given, so that makes sense). However, after the experiment was ran, they ended up with 11.77 grams of the final product. They were told that the 0.35 grams was "lost" as CO2.
Here are the questions.
a) Calculate the mass of NaHCO3 used.
b) Calculate the % yield.
NaHCO3 + HCl = NaCl + H2O + CO2.
In our attempt to answer the questions, we first assumed the 11.77 grams only related to the NaCl + H2O part of the above equation and that the 0.35 grams relates to the CO2. Even though they say it was "lost" that just means it wasn't captured in the beaker - it's not really lost. What's left in the beaker, and what they can weigh, is just the NaCl + H2O which weights 11.77 grams.
Given that, we can use the 0.35 grams of CO2 to calculate the grams of NaHCO3 (using conversions and ratios).
a) The mass of NaHCO3 would be: We think it's 0.6682 g, but need help confirming.
0.35 g CO2 x ( 1 mole CO2 / 44 g CO2 ) x ( 84 g NaHCO3 / 1 mol NaHCO3) = 0.6682 g NaHCO3
b) The % yield would be: Not sure. Need help answering and understanding
We assumed the NaHCO3 is the limiting reactant, because at 10.13 grams of HCl would require 23.6 grams of NaHCO3 in a balanced equation (we got that by 10.13 x 1/36 x 84). Since we only have 1.99 grams of NaHCO3 that limits the HCl to 0.833 grams and the rest is excess.
However, where we got confused is if there is excess NaHCO3, where is it in the right side of the equation? Wouldn't there be a certain amount leftover there?
Homework Statement
The class ran an experiment where they were given 1.99 grams of NaHCO3 and 10.13 grams of HCl to create NaCl, H2O, and CO2. Before starting the experiment, what they were given weighed 12.12 grams in a beaker (which adds up to the 1.99g + 10.13g given, so that makes sense). However, after the experiment was ran, they ended up with 11.77 grams of the final product. They were told that the 0.35 grams was "lost" as CO2.
Here are the questions.
a) Calculate the mass of NaHCO3 used.
b) Calculate the % yield.
Homework Equations
NaHCO3 + HCl = NaCl + H2O + CO2.
The Attempt at a Solution
In our attempt to answer the questions, we first assumed the 11.77 grams only related to the NaCl + H2O part of the above equation and that the 0.35 grams relates to the CO2. Even though they say it was "lost" that just means it wasn't captured in the beaker - it's not really lost. What's left in the beaker, and what they can weigh, is just the NaCl + H2O which weights 11.77 grams.
Given that, we can use the 0.35 grams of CO2 to calculate the grams of NaHCO3 (using conversions and ratios).
a) The mass of NaHCO3 would be: We think it's 0.6682 g, but need help confirming.
0.35 g CO2 x ( 1 mole CO2 / 44 g CO2 ) x ( 84 g NaHCO3 / 1 mol NaHCO3) = 0.6682 g NaHCO3
b) The % yield would be: Not sure. Need help answering and understanding
We assumed the NaHCO3 is the limiting reactant, because at 10.13 grams of HCl would require 23.6 grams of NaHCO3 in a balanced equation (we got that by 10.13 x 1/36 x 84). Since we only have 1.99 grams of NaHCO3 that limits the HCl to 0.833 grams and the rest is excess.
However, where we got confused is if there is excess NaHCO3, where is it in the right side of the equation? Wouldn't there be a certain amount leftover there?