[puremercury]

A mixture of BaCl2 and Ba(NO3)2, with a total mass
2.00 g, is dissolved in water and treated with excess
AgNO3 solution. The precipitate (AgCl) is dried and
found to weigh 0.688 g. How much BaCl2 was in
original mixture?

Molar masses
BaCl2 208.23 g/mol
Ba(NO3)2 261.35 g/mol
AgCl 143.32 g/mol

Can anyone show me the logic to get the answer ? (0.5 g)

[syko sykes]

Use mole to mole ratios to see that it takes .5 mol of BaCl2 to make 1 mol of AgCl. Use the molar mass of AgCl and mass of precipitate to calculate moles formed. Half that for moles of BaCl2 used. Multiply that number by molar mass to get grams.

[puremercury]

K. i got that.  but the thing is how you know how the compound going to react and writing the balance equation in order to get the ratio. that's the part where i got stuck on.

BaCl2+Ba(NO3)2+AgNO3--->2AgCl+BaNO3+BaNO3
?
hum that's odd.

[wereworm73]

You really don't need to bother with the barium nitrate.  The actual reaction is

BaCl₂ + 2 AgNO₃ ---> Ba(NO₃)₂ + 2 AgCl (ppt.)

[syko sykes]

You really don't need to bother with the barium nitrate.  The actual reaction is

BaCl₂ + 2 AgNO₃ ---> Ba(NO₃)₂ + 2 AgCl (ppt.)

Bingo! You know BaNO3 isn't going to react because the anion is the same as the anion of the other reactant, AgNO3. It also won't react with BaCl because the cations are the same. After you know that, it's just a double replacement reaction. Balance and you have mole to mole ratios.

[puremercury]

Thanks