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.
Answer:
Kc = 12.58
Explanation:
Kc = [0.229]^2*[0.687]^6/[0.221]^4*[0.5685]^3
Kc = (0.052441)(0.10513)/(0.002385)(0.18373)
Kc = 0.0005513/0.000438
Kc = 12.58
Hope that helps!!
Answer:
Most nonmetals are solids, but some are gaseous or liquid. All nonmetals are solid unless they bond with a metal.
Explanation:
ANSWER: LOOK IT UP IN YO DICtionary
- williams got dem guns
Answer:
number of moles of NaCl produce = 12 mol
Explanation:
Firstly, we need to write the chemical equation of the reaction and balance it .
Na(s) + Cl2(g) → NaCl(s)
The balanced equation is as follows:
2Na(s) + Cl2(g) → 2NaCl(s)
1 mole(71 g) of chlorine produces 2 moles(117 g) of sodium chloride
6 mole of chlorine gas will produce ? mole of sodium chloride
cross multiply
number of moles of NaCl produce = 6 × 2
number of moles of NaCl produce = 12 moles
number of moles of NaCl produce = 12 mol
