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babunello [35]
3 years ago
14

Specific heat problems for big points. PLEASE help

Chemistry
1 answer:
sesenic [268]3 years ago
3 0

Answer:

though i didnt do that yet. i will google it and be right back to u.

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In this reaction, what roll does the lead (II) nitrate play when 50.0 mL of 0.100M iron (III) chloride are mixed with 50.0 mL of
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Lead(II) nitrate will react with iron(III) chloride to produce the precipitate lead(II) chloride as shown in the balanced reaction
     2FeCl3(aq) + 3Pb(NO3)2(aq) → 2Fe(NO3)3(aq) + 3PbCl2(s)   
Calculating the amount of the precipitate lead(II) chloride each reactant will produce: 
     mol PbCl2 = 0.050L Pb(NO3)2 (0.100mol/1L)(3mol PbCl2/3mol Pb(NO3)2)
                       = 0.00500mol PbCl2
     mol PbCl2 = 0.050L FeCl3 (0.100mol FeCl3/1L)(3mol PbCl2/2mol FeCl3)                                 = 0.00750mol PbCl2
The reactant Pb(NO3)2 produces a lesser amount of the precipitate PbCl2, therefore, the lead(II) nitrate is the limiting reagent for this reaction.
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