From your correct balanced equation, follow the following steps to obtain all of the solutions.
You worked out the balanced equation correctly. The next step is to determine which of the products is the likely precipitate. For purposes of this exercise (no data was given on Ksp) we will assume 100% precipitation.
The third step is to see which is the limiting reagent by converting the masses given to moles, and seeing which of the two is the least moles – that will determine the maximum product available.
The fourth step is to revisit your balanced equation to see how many moles of the precipitate product will be formed, given the limiting reagent quantity. Now that you know how many moles of the precipitate will be formed, you can convert that value into a mass.
The fifth step is to subtract the moles required for the reaction from the limiting reagent to find out how many moles remained in excess. Convert that value to mass for the answer to part b).
Calculate the mass of the precipitate formed when 2.27L of 0.0820M
First calculate the absolute number of moles of each compound.
0.0820M = x/2.27L ==> x = (2.27L) * (0.0820M) = 0.186mol
0.0664M = x/3.06L ==> x = (3.06L) * (0.0664M) = 0.203mol
You must remember to determine the precipitate and balance the reaction! The compounds in solution would just stay there if they didn't react to form an insoluble precipitate.
NOW you can see that 1 mole of
Thus, the amount of precipitate is: 0.186 mol x 233.3 g/mol
How many grams of NaCl are required to precipitate most of the
Na+(aq) + Cl-(aq) +
Net Ionic: Ag+(aq) + Cl-(aq)→AgCl(s)
0.250L x 0.0113 = 0.00283 mol AgNO3
From the Stoichiometry: 1 mole
0.00283 mole x (58.44g/mol NaCl) = 0.165 g
Thus, any MORE NaCl would be an excess reagent.