Liters of CO2 produced by 8.4 g NaHCO3

[za105]

I'm on a ridiculous pace for completing a year's worth of Chemistry... I have two weeks left to finish over a semester's worth, so I'm struggling to learn all the concepts thoroughly and get decent grades. And so I keep getting stuck on problems.

This problem is really giving me a headache:
From the equation 2 NaHCO3 --> Na2CO3 + CO2 + H2O, how many liters of CO2 at STP will be produced by 8.4 g NaHCO3?

a. 1.21 L
b. 1.12 L
c. 12.1 L
d. 11.2 L

I believe the first lengthy equation, "2 NaHCO3 etc." is extraneous. The next steps that I've taken are calculating the total gram mass of NaHCO3, 84 grams, meaning 8.4 grams is 1/10 of NaHCO3. If this is so, I'm figuring there will be 1/10 as many grams of CO2, meaning 44 grams of CO2. 44 grams to liters is, at a conversion factor of 1000 grams per liter, .044 liters... I'm obviously way off, but I'm clueless as to what else I'm supposed to be doing. I can't find any similar problems in all the other practice problems in my textbook or syllabus to give me a clue. I'm homeschooled, so no teachers to talk to.

Help?

 

[mrjeffy321]

I believe the first lengthy equation, "2 NaHCO3 etc." is extraneous.

No, actually it is quite important.

You need to find the molar ratio of moles of NaHCO3 reacted to moles of CO2 produced.

Find the number of moles of NaHCO3 which react, use the mole ratio to find the number of moles of CO2 produced.
If you assume CO2 is an ideal gas, then at STP, 1 mole will have a volume of 22.4 Liters.

 

[Borek]

Read about stoichiometric calculations, that's what you need.

And remember that 1000 g per L is for water, not for gases nor other substances.