Given the reaction 2HI (aq) + CaCO3(s) = CaI2(aq)+CO2(g)+H2O(l) What volume of of CO2 can be produced from 255mL of 3.0M HI and 75.2g of CaCO3 at STP?

75.2g CaCO3/100.09g/mol CaCO3 =.751mol CaCO3

(3M)(.255L)=.765 mol HI

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  1. This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants.

    Your first step is ok.

    Using the coefficients in the balanced equation convert mols CaCO3 to mols CO2.
    Do the same and convert mols HI to mols CO2.
    It is likely that these two values will not be the same; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that number is called the LR.

    Now convert the the LR value in mols to grams CO2. g = mols x molar mass = ?

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  2. Okay so this was my next step

    0.765 mol CaCO3 x 1 mol CO2/2 mol HI = 0.3825mol CO2.

    At STP, 1 mol=22.4L so (22.4)(0.3825mol)= 8.568 L CO2

    Does that look right?

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