Calculating grams of Precipitate

[hviz]

Homework Statement

If 70.0 mL of 0.150 M CaCl2 is added to 15.0 mL of 0.100 M AgNO3, what is the mass in grams of AgCl precipitate?

Homework Equations

Limiting reagants.

The Attempt at a Solution

So I tried to find the limiting reagent, but that's where I'm stuck. I'm using examples from my textbook and notes, but everything is in g instead of mL/L. So it's quite confusing. I don't even know how to start.
I tried:
(.15M CaCl2)(.07L CaCl2)(1 mol CaCl2/2 mol AgNO3)
And that's about as far as I got, because I didn't even know if I was doing the right steps.

Any help is very appreciated.

 

[symbolipoint]

To be more sure, first compare the number of moles of CaCl2 to the number of moles of AgNO3; then, use the ratio information of 1 mole Calcuim Chloride to 2 moles Silver Nitrate to determine which compound is the limiting reactant.

Moles calcium chloride available = 0.07L * 0.15 M = ( )
moles silver nitrate available = 0.015L * 0.1 M = ( )

Which will one will still have a portion unreacted after mixing?

 

[Blueshift5]

To find the mass of the precipitate, we will first need to determine the limiting reagent. This is the reactant that will be completely used up in the reaction, and it will determine the amount of product that can be formed. To do this, we can use the mole ratio between CaCl2 and AgNO3, which is 1:2 according to the balanced chemical equation.

First, we need to convert the volumes of the solutions to moles. Using the given concentrations, we can calculate the moles of CaCl2 as (0.150 mol/L)(0.070 L) = 0.0105 mol. Similarly, the moles of AgNO3 can be calculated as (0.100 mol/L)(0.015 L) = 0.0015 mol.

Since the mole ratio between CaCl2 and AgNO3 is 1:2, we can see that 0.0105 mol of CaCl2 is equivalent to 0.00525 mol of AgNO3. This means that AgNO3 is the limiting reagent, as it will be completely used up before all of the CaCl2 is consumed.

To calculate the mass of the precipitate, we can use the molar mass of AgCl, which is 143.32 g/mol. We know that 0.0015 mol of AgNO3 will form 0.0015 mol of AgCl, so we can use this to calculate the mass of AgCl as (0.0015 mol)(143.32 g/mol) = 0.21498 g.

Therefore, the mass of AgCl precipitate formed in this reaction is 0.215 g. It is always important to check your units and make sure they are consistent throughout the calculation. I hope this helps with your understanding.