chem question

[mR LaZy]

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how much CO2 is produced @stp from 100ml of solution made by mixing 100g of KHCO3 into enough solution to form 500ml?

rxn eq = HBr+KHCO3 -> KBr +h2o + co2

thanks

 

[mR LaZy]

i think i solved it

100g x 1mole khco3/95g = 1.05 moles/.5L = 2 x .1L x 22.4L@STP = 4.48L of CO2

anyone have any suggestions?

 

[ksmi117]

looks good to me 🙂

 

[w0lf137]

i think i solved it

100g x 1mole khco3/95g = 1.05 moles/.5L = 2 x .1L x 22.4L@STP = 4.48L of CO2

anyone have any suggestions?

To save some time (skipping the concentration part of the equation), because 1 mole of khco3 will break down to 1 mole of CO2 = 22.4L. That 22.4L is reduced by a 1/5 when the dude takes 100ml out of the 500 ml. In other words the 1 mole of CO2 in the 500 mL is reduced to a certain # of moles in the 100mL. Going from 500 ---> 100 is 1/5. So reducing the 1 mole or 22.4L (at stp) by a 5th is: 22.4/5 is around 5L.