# chemistry

Determine the molarity of nitrate ions in a solution prepared by mixing 25.0 mL of 0.50 M Fe(NO3)3 and 35.0 mL of 1.00 M Mg(NO3)2 Please explain fully do not know where to start?

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moles Fe(NO3)3 = M x L = ??
moles NO3^- will be 3x that.

moles Mg(NO3)2 = M x L = ??
moles NO3^- will be 2x that.

Then M NO3^- = total moles NO3^-/total volume in L.

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2. (0.50)(25.0) = 12.5

(1.00)(35.0) = 35

47.5/60 = 0.79M is my answer but my book says

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3. You didn't follow any of my instructions. I have typed my bolded instructions below each of your lines.
0.50)(25.0) = 12.5

moles Fe(NO3)3 = M x L = ??
moles NO3^- will be 3x that.
M x L = 0.50M x 0.025 L(not 25 mL) = 0.0125 moles Fe(NO3)3
Then nitrate is 3x that or 0.0125*3 = 0.03750 moles nitrate ion

(1.00)(35.0) = 35
moles = M x L = 1.00M x 0.0350 L (not mL) = 0.0350 moles Mg(NO3)2.
Then nitrate is twice that or
2*0.0350 = 0.0700 M in nitrate.

47.5/60 = 0.79M is my answer but my book says
M nitrate = total moles/total volume in L. (0.0375+0.0700)/0.06L = 1.79 M in nitrate.

So it is

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4. so I times 0.0125 by 3 b/c of the 3 at the end of the bracket?

and I times 0.0350 by 2 b/c of the 2 at the end of the bracket?

but why because when i'm doing my molar mass i do include that last # in the molar mass so why do we times it by that #?

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