Quote:
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Originally Posted by sdekivit
Quote:
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Originally Posted by redXI
hey are you sure the answer is 0,21 g?? I tried for about 10 times but asnwer is much bigger than that.  |
i'm 100% sure about that. 1,5 mmol AgCl will be formed and thus 215 mg. |
But we need to balance the equation right?
So it's like this:
2 AgNO3 + CaCl2 --> 2 AgCl + Ca(NO3)2
1,5 mole 4,5 mole
then.... we divide the 1,5 mole with 2 = 0,75 mole
and then finding the mass of AgCl = 2/2 x 0,75 x 143,35 = 107, 5 mg.