good chemistry question

[pauly11235]

Hi, I have a problem here:

How many grams of calcium chloride are needed to prepare 72g of silver chloride according to the following equation?

CaCl2(aq) + 2AgNO3(aq) --> Ca(NO3)2(aq) + 2AgCl

no idea how to do it but i beilive the answer is 28.0g. can someone please show me how to do it.

Thanks again!

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[nixon13]

Hi, I have a problem here:

How many grams of calcium chloride are needed to prepare 72g of silver chloride according to the following equation?

CaCl2(aq) + 2AgNO3(aq) --> Ca(NO3)2(aq) + 2AgCl

no idea how to do it but i beilive the answer is 28.0g. can someone please show me how to do it.

Thanks again!

72g AgCl x (1 mol/MM AgCl) x (1molCaCl2/2molAgCl) x (MMCaCl2/1molCaCl2) = your answer.

 

[jdog10]

Hi, I have a problem here:

How many grams of calcium chloride are needed to prepare 72g of silver chloride according to the following equation?

CaCl2(aq) + 2AgNO3(aq) --> Ca(NO3)2(aq) + 2AgCl

no idea how to do it but i beilive the answer is 28.0g. can someone please show me how to do it.

Thanks again!

Divide the grams of AgCl by its molar mass, then multiply that number (which is moles of AgCl) by the mole ratio which is 1 mole CaCl2 / 2 moles AgCl (so 1/2) and finally multiply that number (which is moles of CaCl2) by the molar mass of CaCl2. That should get you the correct answer although I don't have a periodic table by me to check.

(72 g AgCl / MM AgCl) x (1 mole CaCl2 / 2 moles AgCl) x (MM CaCl2) = grams CaCl2

 

[topdent1]

Hi, I have a problem here:

How many grams of calcium chloride are needed to prepare 72g of silver chloride according to the following equation?

CaCl2(aq) + 2AgNO3(aq) --> Ca(NO3)2(aq) + 2AgCl

no idea how to do it but i beilive the answer is 28.0g. can someone please show me how to do it.

Thanks again!

72g AgCl x (1 mol/MM AgCl) x (1molCaCl2/2molAgCl) x (MMCaCl2/1molCaCl2) = your answer.

Here's how you do it.

1. You first convert 72 g of AgCl in to the number of moles. 72/143 = about .5 moles.
2. Now compare the mole ratios of CaCl2 and AgCl. 1:2. That means that 1/4 moles of CaCl2 will give .5 moles of CaCl2 right? Yes.
3. Now just multiply 1/4 moles of CaCl2 by the MW of CaCl2, and that should give you your answer.

 

[rdhdds1]

Hi, I have a problem here:

How many grams of calcium chloride are needed to prepare 72g of silver chloride according to the following equation?

CaCl2(aq) + 2AgNO3(aq) --> Ca(NO3)2(aq) + 2AgCl

no idea how to do it but i beilive the answer is 28.0g. can someone please show me how to do it.

Thanks again!

[(72 g AgCl)(1 mole AgCl)/143 g AgCl][(1 mole CaCl2)/2 moles AgCl][(110 g CaCl2)/1 mole CaCl2] = 27.69 g CaCl2