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1) If 740. mL of 0.06622 M aqueous NaI and 370. mL of 0.07151 M aqueous Pb(NO3)2 are reacted stoichiometrically according to the equation, what mass (g) of NaI remained?

Pb(NO3)2(aq) + 2 NaI(aq) → PbI2(s) + 2 NaNO3(aq)

2) If 180 mL of 0.581 M aqueous HCl and 690 mL of 0.0696 M aqueous CO32- are reacted, how many mol of liquid H2O is produced.

Na2CO3(aq) + 2 HCl(aq) → CO2(g) + 2 NaCl(aq) + H2O(l)

  • college chemistry -

    Both problems are limiting reagent problems. For #2,
    Convert mL and M to moles. moles = M x L.
    Using the coefficients in the balanced equation, convert moles HCl to moles H2O.
    Using the same procedure, convert moles Na2CO3 to moles H2O.
    You will likely obtain different answers for moles H2O; in limiting reagent problems the smaller value is ALWAYS the correct one and the reagent producing that value is the limiting reagent.

    For #1, use the above procedure to identify the limiting reagent. Then, using the coefficients in the balanced equation, convert moles of the non-limiting reagent to moles of the limiting reagent, convert moles of the limiting reagent to grams (g = moles x molar mass) and subtract from the grams originally present. The result is g NaI remaining. From the way the problem is presented, I assume Pb(NO3)2 is the limiting reagent and NaI is the non-limiting reagent.
    Post your work if you get stuck.

  • college chemistry -

    thank youu!

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