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)
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.
thank youu!
To determine the mass of NaI remaining in the first reaction and the number of moles of liquid H2O produced in the second reaction, we can use the principles of stoichiometry. Here's how you can solve each problem step by step:
1) Mass of NaI remaining:
- Start by calculating the number of moles of NaI and Pb(NO3)2 present in the given volumes of their respective solutions.
- First, use the formula: moles = concentration (M) x volume (L).
- For NaI: moles of NaI = 0.06622 M x 0.740 L.
- For Pb(NO3)2: moles of Pb(NO3)2 = 0.07151 M x 0.370 L.
- Since the stoichiometric ratio between NaI and Pb(NO3)2 is 1:1, the moles of NaI is the limiting reagent.
- Now, calculate the moles of NaI reacted by multiplying it by the stoichiometric coefficient of NaI: moles of NaI reacted = moles of NaI x 1.
- The moles of NaI remaining is the initial moles of NaI minus the moles of NaI reacted.
- Finally, calculate the mass of NaI remaining using the formula: mass = moles x molar mass of NaI.
2) Moles of liquid H2O produced:
- Begin by calculating the number of moles of HCl and CO32- in the given volumes of their respective solutions.
- Use the same formula: moles = concentration (M) x volume (L).
- For HCl: moles of HCl = 0.581 M x 0.180 L.
- For CO32-: moles of CO32- = 0.0696 M x 0.690 L.
- Based on the stoichiometric coefficients, the mole ratio between HCl and H2O is 2:1.
- Therefore, the number of moles of HCl will determine the number of moles of H2O produced.
- Multiply the moles of HCl by 2 to find the moles of H2O produced.
Remember to always double-check your calculations and units.