# Question #c4072

Mar 23, 2016

From your correct balanced equation, follow the following steps to obtain all of the solutions.

#### Explanation:

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).

Examples follow:
Calculate the mass of the precipitate formed when 2.27L of 0.0820M $B a {\left(O H\right)}_{2}$ are mixed with 3.06L of 0.0664M $N {a}_{2} S {O}_{4.}$

First calculate the absolute number of moles of each compound.
0.0820M = x/2.27L ==> x = (2.27L) * (0.0820M) = 0.186mol $B a {\left(O H\right)}_{2}$
0.0664M = x/3.06L ==> x = (3.06L) * (0.0664M) = 0.203mol $N {a}_{2} S {O}_{4}$

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.

$B a {\left(O H\right)}_{2} + N {a}_{2} S {O}_{4} - \to B a S {O}_{4} \left(s\right) + 2 N {a}^{+ 1} + 2 O {H}^{- 1}$ (ions in solution)

NOW you can see that 1 mole of $B a {\left(O H\right)}_{2}$will react with one mole of $N {a}_{2} S {O}_{4}$ to form 1 mole of $B a S {O}_{4.}$ However, we only have 0.186 mol of $B a {\left(O H\right)}_{2}$ and 0.203 mol of $N {a}_{2} S {O}_{4} ,$ as we calculated. So, we will end up with 0.186 mole of $B a S {O}_{4} ,$as once the $B a {\left(O H\right)}_{2}$is used up, there will be no more precipitate formed. This is also called the limiting reagent of a reaction, and will be on the quizzes and exams.

Thus, the amount of precipitate is: 0.186 mol x 233.3 g/mol $B a S {O}_{4}$ = 43.4 g

How many grams of NaCl are required to precipitate most of the $A {g}^{+}$ ions from $2.50 x {10}^{2}$ ml of 0.0113 M $A g N {O}_{3}$ solution?

Na+(aq) + Cl-(aq) + $A g N {O}_{3}$ (aq) ----> NaNO3(aq) + AgCl(s)
Net Ionic: Ag+(aq) + Cl-(aq)→AgCl(s)

Given: $A g N {O}_{3}$= 0.0113 moles/L (Molarity) and 2.50 x 102 ml
0.250L x 0.0113 = 0.00283 mol AgNO3

From the Stoichiometry: 1 mole $A g N {O}_{3}$ reacts with 1 mole NaCl so the number reacted is equal to: 0.00283 mole.
0.00283 mole x (58.44g/mol NaCl) = 0.165 g
Thus, any MORE NaCl would be an excess reagent.