2NaHCO3 + heat ==> Na2CO3 + H2O + CO2
I would set this up like this.
Let X = mass NaHCO3
and Y = mass NaCl
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g NaHCO3 + g NaCl = 1.124 and
g Na2CO3 + g NaCl = 1.124 - 0.270
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X + Y = 1.124
X(1 mol Na2CO3/1 mols NaHCO3) + Y = 0.854
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X + Y = 1.124
X(106/2*84) + Y = 0.854
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X + Y = 1.124
0.631X + Y = 0.854
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Solve these two equation simultaneously for X. That will give you grams NaHCO3. You should look up the molar masses of Na2CO3 and NaHCO3 and use a little more accuracy that my rounded numbers of 106 and 84.
After you find g NaHCO3, then
%NaHCO3 = (mass NaHCO3/mass sample)*100 = ??
My numbers are 0.732 g NaHCO3 for 65.1% which should be close.
Heating 1.124g of an unknown NaHCO3-NaCl mixture resulted ina mass loss of 0.270g.
1) Calculate the mass of NaHCO3 present in the original mixture.
2) Calculate the mass percent of NaHCO3 in the uknown mixture
2 answers
thank you