ManzRDH

Dinene RDH, BA
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How many grams of CaCl are need to prepare 72 grams of silver chloride according to the following:

CaCl(aq) + 2 AgNO2(aq)------ Ca(No3)2 (aq) + 2 AgCl

a-27 g
b- 28g
c- 29g
d-30 g
e-none of the above
 

Dentist 2 be

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72 g * (1 mol CaCl2/ 2 mol AgCl) * (110g CaCl2 / 144 g AgCl) = 27.5 g of CaCl2 are needed to produce 72g of AgCl.

I would say the correct answer is E, none of the above. Unless they want you to estimate that 27.5g is 28g, but I doubt it since all the answer choices a re about 1g off
 

PEN15

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ManzRDH said:
How many grams of CaCl are need to prepare 72 grams of silver chloride according to the following:

CaCl2(aq) + 2 AgNO2(aq)------ Ca(No3)2 (aq) + 2 AgCl

a-27 g
b- 28g
c- 29g
d-30 g
e-none of the above
I usually do everything in moles:
1. convert 72g AgCl into moles = 72g/(144g/mol) = 0.5 mol
2. since there is one mole of calcium chloride needed to produce 2 moles of silver chloride, you just divide by 2 to give your moles of calcium chloride. 0.25 mol CaCl2 (if the question was the other way around you would multiply by 2)
3. molar mass of CaCl2 = 110g/mol
4. in order to just get grams, you need to get rid of the mol so you multiply 110g/mol * 0.25mol = 27.5 g of CaCl2

I find that when doing calculations like these, keeping your units help. This is just another explanation to the problem. The previous poster's method is perfectly feasible.