If 270. mL of 0.08089 M aqueous Pb(NO3)2 and 710. mL of 0.05493 M aqueous Na+ are reacted stoichiometrically according to the equation, what mass (g) of Na+ remained?
Pb(NO3)2(aq) + 2 NaI(aq) → PbI2(s) + 2 NaNO3(aq)
You see how that works from the CaCO3 and H2SO4 problem just above so I think you can do this on your own. ust follow the steps. Post your work if you get stuck.
To solve this problem, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed, determining the maximum amount of product that can be formed.
We can start by calculating the number of moles of each reactant using the given volumes and concentrations. We'll use the equation:
moles = concentration (M) x volume (L)
For Pb(NO3)2:
moles of Pb(NO3)2 = 0.08089 M x 0.270 L = 0.02188 mol
For Na+:
moles of Na+ = 0.05493 M x 0.710 L = 0.0390 mol
Now, let's compare the moles of each reactant to their stoichiometric coefficients in the balanced equation. In this case, the stoichiometric coefficient of Pb(NO3)2 is 1, and the stoichiometric coefficient of Na+ is 2.
Since the stoichiometric ratio of Pb(NO3)2 to Na+ is 1:2, we need twice as many moles of Na+ as Pb(NO3)2 to react completely. However, we only have 0.02188 mol of Pb(NO3)2 and 0.0390 mol of Na+. This means that Pb(NO3)2 is the limiting reactant because we have fewer moles of Pb(NO3)2 than Na+ needed.
To find the mass of Na+ remaining, we'll need to determine how many moles of Na+ react with the available moles of Pb(NO3)2. Since the stoichiometry of the balanced equation tells us that 2 moles of Na+ react with 1 mole of Pb(NO3)2, and we have 0.02188 mol of Pb(NO3)2, we can calculate the moles of Na+ required:
moles of Na+ required = 2 x 0.02188 mol = 0.04376 mol
Now, to find the moles of Na+ remaining, we can subtract the moles of Na+ required from the initial moles of Na+:
moles of Na+ remaining = 0.0390 mol - 0.04376 mol = -0.00476 mol
The negative value indicates that all the Na+ has been consumed, and there is no excess Na+ remaining. So, the mass of Na+ remaining is 0 grams.
Therefore, there is no mass of Na+ remaining after the reaction.