How many grams of Ag2CO3 will precipitate when excess Na2CO3 solution is added to 68.0 mL of 0.642 M AgNO3 solution?

6.07 grams of Ag2CO3 will precipitate out.

Explanation:

STEP ONE:

As with many other problems in chemistry, a great first step is to write out a balanced chemical equation so that we know what we're dealing with before we actually start working.

In this case, we can see that they are adding a solution of Na2CO3 to a solution of AgNO3. This will give us the following balanced chemical equation:

2AgNO3  +  Na2CO3  >  2NaNO3  +  Ag2CO3

What we understand from this equation is that when TWO moles of AgNO3 are reacted with ONE mole of Na2CO3, we get a product of TWO moles of NaNO3 and ONE mole of the product we're interested in; Ag2CO3.

STEP TWO:

It is very important that you understand that every time a teacher gives you the volume AND the molarity of something, they are giving you the moles of that something. This works because we are given 68.0 mL and molarity happens to have the units of Mol/L, or Moles of the substance divided by Liters of solution. In general, when Liters of solution aren't given in the problem, we can just assume it's one liter because it's a nice number to work with.

First, we'll convert 68 mL to Liters using dimensional analysis. Then, we'll multiply that by the molarity.

This is what that looks like:

(68.0 mL) * (1 L / 1000 mL)  =   0.068 L

(0.068 L) *  (0.642 moles / 1 Liter)  =   0.044 mol AgNO3

Liters multiplied by Moles / Liter cancels out the Liters and we're left with just moles, which is 0.044 mol AgNO3 in this case.

STEP THREE:

Now that we have converted to the moles of the solution we want, AgNO3, we can take a look at what this problem is asking us to do.

Here, we need to do some Stoichiometry to convert from moles of AgNO3 to moles of Ag2CO3. That's why we got moles in the first place is that stoichiometry is always done using moles.

2AgNO3  +  Na2CO3  >  2NaNO3  +  Ag2CO3

From the balanced chemical equation, we can see that whenever two moles of AgNO3 react, we get a product of one moles of Ag2CO3. This is very important to note.

Here's the Stoichiometry:

(0.044 mol AgNO3)  *  (1 mol Ag2CO3 / 2 mol AgNO3)  =   0.022 mol Ag2CO3

So, moles of AgNO3 cancel out and we're left with moles of 0.022 mol Ag2CO3.

STEP FOUR:

Since the question asks for grams, we aren't quite done yet. We need to do a mole to gram conversion. First, we'll go to our periodic table and add up the constituents of our compound of Ag2CO3 to get the molar mass.

(2 moles Ag)  215.74g  +   (1 mol C) 12.011g  +  (3 moles O) 47.99g

= 275.74 g per mole of Ag2CO3

Now we can simply use the moles of Ag2CO3 we calculated from before and multiply by the molar mass.

(0.022 mol Ag2CO3)  *   (275.74g / 1 mol Ag2CO3)

= 6.07 grams Ag2CO3

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