Answer:
21 mL of NaOH is required.
Explanation:
Balanced reaction: 
Number of moles of HBr in 11.0 mL of 0.30 M HBr solution
=
moles = 0.0033 moles
Let's say V mL of 0.16 M NaOH solution is required to reach equivalence point.
So, number of moles of NaOH in V mL of 0.16 M NaOH solution
=
moles = 0.00016V moles
According to balanced equation-
1 mol of HBr is neutralized by 1 mol of NaOH
So, 0.0033 moles of HBr are neutralized by 0.0033 moles of NaOH
Hence, 

So, 21 mL of NaOH is required.
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 would be A)the atom
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