daniaridha Posted October 5, 2014 Report Share Posted October 5, 2014 I would REALLY appreciate some full explanation for this question. The thermal decomposition of sodium hydrogen carbonate, NaHCo3, results in a 73.8% yield of sodium carbonate, Na2Co3. 2NaHCo3=Na2Co3+H2O+Co2 If a 1.68 g sample of sodium hydrogen carbonate is heated, calculate the mass, in g, of sodium carbonate produced. Reply Link to post Share on other sites More sharing options...
emzinger Posted October 14, 2014 Report Share Posted October 14, 2014 you would have to use moles, its pretty tedious Reply Link to post Share on other sites More sharing options...
CriCri Posted October 14, 2014 Report Share Posted October 14, 2014 You have to work with moles. If two moles of NaHCO3 react to form 1 of Na2CO3 then u write a ratio. First you have to calculate the mass of both these compounds, and this will be one mole. For NaHCO3 you have to calculate two moles. So you write if x grams of NaHCO3 (make sure you calculate the mass of 2 moles of it) react with y moles of Na2CO3 (make sure the mass is one mole) then 1.68g of NaHCO3 will react to produce y grams of Na2CO3.Hope this helps!! I would REALLY appreciate some full explanation for this question. The thermal decomposition of sodium hydrogen carbonate, NaHCo3, results in a 73.8% yield of sodium carbonate, Na2Co3. 2NaHCo3=Na2Co3+H2O+Co2 If a 1.68 g sample of sodium hydrogen carbonate is heated, calculate the mass, in g, of sodium carbonate produced. Reply Link to post Share on other sites More sharing options...
Chemstryandbio Posted October 14, 2014 Report Share Posted October 14, 2014 1) Find RFM of NaHCO3.2) moles= mass/RFM moles=1.68/168.023) moles= 0.00999884) moles/73.8, Ansx100.5) 0.0135 mol6) Calculate RFM of Na2CO3x0.02357)105.99x0.0235= 1.44g Reply Link to post Share on other sites More sharing options...
Turquoise_96 Posted October 22, 2014 Report Share Posted October 22, 2014 n(NaHCO3)=1.68g/84 [this is the Mr(NaHCO3)] = 0.02mol according to the molar ratio of the equation, 2 moles of NaHCO3 yields 1 mol of Na2CO3, hence:n(Na2CO3)= 0.02mol/2 = 0.01 moltheoretical yield of Na2CO3= n(Na2CO3) x Mr (Na2CO3)= 0.01mol x 106 = 1.06gActual yield in grams = (73.8/100) x 1.06g = 0.782g. Is this the correct answer? Reply Link to post Share on other sites More sharing options...
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