ilovemedi

5+ Year Member
Nov 21, 2011
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Question: For the unbalanced rxn HBR +KHCO3 --> KBR +water +Co2, how many liters of CO2 are produced at STP from 100mL of a solution made by mixing 100 g KHCO3 (100g/mole) into enough solved to form 500 mL of solution?

A)1.12L
B)2.24L
C)4.48L
D) 5.6L

Answer (highlight)': C

Problem: TOO MANY NUMBERS! I got confused SO easily. Where do I start, if this was a question on the MCAT? I'd freeze. All I can get to is converting 100g/mole to CO2 g/mole then 22.4L=1 mole. HELP! :)
 
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ilovemedi

5+ Year Member
Nov 21, 2011
327
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Ok, is this the right thought process: So you have 100g=1mole Co2, which also equals 1 mol CO2 (no idea why it says unbalanced, because it really is?). So 1mol Co2*22.4L/mol = 22.4Liters.. BUT HEY - we're only sampling 100ml/500ml so it's 1/5*22.4 = B
 
Aug 5, 2012
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First, you need to check that the reaction is balanced. Fortunately, it already is but I want to explicate the ratios:

1 mol HBr + 1 mol KHCO3 -> 1 mol KBr + 1 mol H2O + 1 mol CO2

Now, we have to pick apart the prompt. It says that 100 mL of a solution made with 100 g KHCO3 and 500 mL solvent was used as (what I'm assuming is) the limiting reagent in this reaction.

First, we need to find out how many moles KHCO3 are in that 500 mL solution:

100 g KHCO3 * (1 mol KHCO3 / 100 g KHCO3) = 1 mol KHCO3

Next, we need to find out the molarity of that 500 mL solution:

1 mol KHCO3 / 500 mL = 1 mol KHCO3 / 0.5 L = 2 M KHCO3 = 2 mol KHCO3 / 1 L

Now, only 100 mL of this solution was used in the reaction so we need to find out how many moles of KHCO3 are involved:

(2 mol KHCO3 / 1 L) * [100 mL * (1 L / 1000 mL)] = (2 mol KHCO3 / 1 L) * (.100 L) = .2 mol KHCO3

Since only .2 mol KHCO3 were involved in the reaction, and an equivalent number of moles of CO2 are produced per mole KHCO3, only .2 mol CO2 are produced. Since we know that 1 mol of gas has a volume of 22.4 L, we multiply that volume by .2 mol:

.2 mol KHCO3 * (1 mol CO2 / 1 mol KHCO3) * (22.4 L CO2 / 1 mol CO2) = 4.48 L CO2