Asked by eng
                 When a reaction mixture with a total volume of 1720 mL that contains 0.605 L of gaseous CO2 measured at STP dissolved in 940 mL of water was stoichiometrically produced as per the balanced equation, how many mol of solid CaCO3 were required?
H2SO4(aq) + CaCO3(s) → CO2(g) + CaSO4(s) + H2O(l)
.605L/22.4L=moles CO2
mol CaCO3=molCO2
            
        H2SO4(aq) + CaCO3(s) → CO2(g) + CaSO4(s) + H2O(l)
.605L/22.4L=moles CO2
mol CaCO3=molCO2
Answers
                    Answered by
            DrBob222
            
    I'm not quite sure how to interpret the problem. If it is 0.605 L CO2 (and you are taking the entire amount of CO2) then your set up is correct. However, if you have CO2 dissolved in 940 mL and you are taking 605 mL of that solution, .....I don't know. Perhaps that scenario doesn't make sense. 
    
                    Answered by
            eng
            
    how would you do it differently then
    
                    Answered by
            DrBob222
            
    I don't know that I can offer another solution. It may be that all of those different mL and L are thre just to confuse you. And they succeeded for me. The method you have is ok, I think, if they interpret it the same way we did.
    
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