How to Calculate Grams of Precipitate in a Chemical Reaction?

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SUMMARY

The discussion focuses on calculating the mass of AgCl precipitate formed when 70.0 mL of 0.150 M CaCl2 is mixed with 15.0 mL of 0.100 M AgNO3. The key steps involve determining the limiting reagent by calculating the moles of each reactant. The moles of CaCl2 are calculated as 0.0105 moles, while the moles of AgNO3 are 0.0015 moles. Since the reaction ratio is 1:2, AgNO3 is the limiting reagent, leading to the formation of 0.0015 moles of AgCl precipitate, which corresponds to a mass of 0.1575 grams.

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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.
 
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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?
 

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