#1. You have the equation.
2. Convert 199 mL of 1.96 M Hg(NO3)2 to moles. moles = M x L.
3. Using the coefficients in the balanced equation, convert moles Hg(NO3)2 to moles NaI.
4. Now convert moles NaI to mL.
M = moles/L. YOu have M and moles, calculate L and convert to mL.
The second problem is done the same way.
I need help with these problems ASAP:
I need help setting these problems up: pleas!!
(1)Calculate the volume of 1.87 M NaI that would be needed to precipitate all of the Hg+2 ion from 199 mL of a 1.96 M Hg(NO3)2. The equation for the reaction is
2NaI (aq) + Hg(NO3)2 (aq) ---> HgI2 (2) +2NaNO3 (aq)
(2)Calculate the volume of 1.91M sulfuric acid that would be needed to neutralize 42.9 mL of a 1.24 M aqueous ammonia solution. The equation for the reaction is:
H2SO4 (aq) +2NH3 (aq) ---> 2NH4+ (aq) + SO4-2 (aq)
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