finding the mass of iron and Fe(NO3)3 given mass of precipitate

[kickitleft7]

Hi I need help with two chemistry problems for my summer hmwrk.  The first one states: A mixture contains only NaCl and Fe(NO3)3. A .456 gram sample of the mixture is dissolved in water and an excess of NaOH is added, producing a precipitate of Fe(OH)3. The precipitate is dried and filtered, and its mass is determined to be .107 g  ---What is the mass of iron in the sample and what is the mass of Fe(NO3)3 om the sample?

My second question i need help with is : A .500L sample of H2SO4 was analyzed by taking a 100.0 mL portion and adding 50.0 mL of .213 M NaOH.  After the reaction occured, an excess of OH ions remained in solution.  This excess base required 13.21 mL of .103 M HCL to neutralize it. Calculate the molarity of the original sulfuric acid

Thanks Again i appreciate anyone's help

[valdorod]

There are several ways to approach these problems.
For problem 1 you can
1) find the percent Fe in Fe(OH)3 and multiply times 0.107 g to obtain the mass of iron in the sample

       a.     (molas mass of Fe)/(molar mass of Fe(NO3)3) *100 = % Fe

2) use the mass of Fe obtained to find the moles of Fe, then from moles of Fe find the moles of Fe(NO3)3 and then mass of Fe(NO3)3.

       a.     mass of Fe to moles of Fe divide by the molar mass of Fe

       b.     there is only one mole of Fe in Fe(NO3)3 so the moles of Fe = moles of Fe(NO3)3

       c.     from moles of Fe(NO3)3 to mass of Fe(NO3)3 multiply times the molar mass of Fe(NO3)3

For problem two you need to set up the balanced equation for the reaction between sulfuric acid and sodium hydroxide.  It is an acid base reaction.

the total moles of NaOH is equal to the moles that reacted with H2SO4 + moles reacted with HCl

thus;  mol H2SO4 = mol NaOH - mol HCl

divid by the volume of H2SO4 to obtain the molarity.