Chemistry Help

[Vivilacqua1]

Alright, so I know you all have seen way too many questions on the topic of chemistry, but here's one more for you all:

 

So we did a lab and I am completely lost on what to do next. We have to find the chemical equation that represents the decomposition of sodium bicarbonate (NaHCO3) and we are given a choice between:

 

2NaHCO3 --> Na2CO3 + CO+ H2O

or           

2NaHCO3 --> Na2O + 2CO2+ H2O

 

 

I am looking at the solids, the first product, Na2O or Na2CO3, and fount that Na2O is .023 moles and Na2CO3 is .012 moles from our 1.42g of precipitate. Any ideas of what to do from here? Do you need any more information?

[don_svetlio]

both are wrong

Use upper and lower case T_T

[silentmelodies]

This topic reminds me why i almost failed chemistry in elementary. Barely passed. 

[don_svetlio]

2NaHCO₃ --> Na₂O + 2CO₂ + H₂O

[Marziman]

Alright, so I know you all have seen way too many questions on the topic of chemistry, but here's one more for you all:

 

So we did a lab and I am completely lost on what to do next. We have to find the chemical equation that represents the decomposition of sodium bicarbonate (NaHCO3) and we are given a choice between:

 

2NaHCO3 --> Na2CO3 + CO2 + H2O

or           

2NaHCO3 --> Na2O + 2CO2 + H2O

 

 

I am looking at the solids, the first product, Na2O or Na2CO3, and fount that Na2O is .023 moles and Na2CO3 is .012 moles from our 1.42g of precipitate. Any ideas of what to do from here? Do you need any more information?

You should be using Open Study or Peer Answer.

[Vivilacqua1]

2NaHCO₃ --> Na₂O + 2CO₂ + H₂O

how were you able to get the subscripts? 

[don_svetlio]

how were you able to get the subscripts? 

The forum has it built in - above the text field

[Vivilacqua1]

The forum has it built in - above the text field

Thanks, edited op

[Sakkura]

Well the second one is what happens, but to derive that from your experimental results you should calculate the amount of substance of the sodium bicarbonate you started with. Then you can compare that number to the numbers you calculated for sodium carbonate or sodium oxide as product.