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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.

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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.

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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 mol

theoretical yield of Na2CO3= n(Na2CO3) x Mr (Na2CO3)= 0.01mol x 106 = 1.06g

Actual yield in grams = (73.8/100) x 1.06g = 0.782g. 

 

Is this the correct answer?

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