If 229 g of Pb(NO3)2 and 190 mL of 1.68 M aqueous NaI are reacted stoichiometrically according to the balanced equation, how many grams of Pb(NO3)2 remain? Round your answer to 3 significant figures.
Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)
Sammy, Todd, John, whomever. This problem is almost the same as the last one. You try it on your own. Post your work if you stuck.
To find out how many grams of Pb(NO3)2 remain after the reaction, we need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed and determines the amount of product formed.
Step 1: Calculate the number of moles of Pb(NO3)2 and NaI.
Molar mass of Pb(NO3)2 = (207.2 g/mol + 3 * 16.0 g/mol) = 331.2 g/mol
Moles of Pb(NO3)2 = 229 g / 331.2 g/mol = 0.6917 mol
Molarity of NaI = 1.68 M
Volume of NaI = 190 mL = 190/1000 L = 0.190 L
Moles of NaI = Molarity * Volume = 1.68 M * 0.190 L = 0.3192 mol
Step 2: Determine the stoichiometric ratio between Pb(NO3)2 and NaI.
From the balanced equation, we can see that the stoichiometric ratio between Pb(NO3)2 and NaI is 1:2. This means that for every 1 mole of Pb(NO3)2, we need 2 moles of NaI.
Step 3: Compare the moles of Pb(NO3)2 and NaI to find the limiting reactant.
Since the stoichiometric ratio between Pb(NO3)2 and NaI is 1:2, we need 2 times the moles of Pb(NO3)2 to react completely with NaI.
Multiplying the moles of Pb(NO3)2 by 2 gives 0.6917 mol * 2 = 1.3834 mol.
Since 0.3192 mol of NaI is less than 1.3834 mol, NaI is the limiting reactant.
Step 4: Find the moles of Pb(NO3)2 remaining.
Since we have excess Pb(NO3)2, the moles of Pb(NO3)2 remaining will be equal to the initial moles minus the moles reacted.
Moles of Pb(NO3)2 remaining = 0.6917 mol - 2 * 0.3192 mol = 0.0533 mol
Step 5: Calculate the mass of Pb(NO3)2 remaining.
Mass of Pb(NO3)2 remaining = Moles of Pb(NO3)2 remaining * Molar mass of Pb(NO3)2
Mass of Pb(NO3)2 remaining = 0.0533 mol * 331.2 g/mol = 17.6476 g
Rounding to 3 significant figures, the mass of Pb(NO3)2 remaining is 17.6 g.