Question

16. What is the limiting reactant when 1.50 g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advanced batteries, according to the following unbalanced equation? Li+N, — LizN 17. What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO, requires 20.22 mL of the AgNO3 solution to reach the end point? AgNO3(aq) + NaCl(aq) AgCl(s) +NaNO (aq) 18. In a common medical laboratory determination of the concentration of free chloride ion in blood serum, a serum sample is titrated with a Hg(NO3)2 solution. 2CF (aq) + Hg(NO3)2(aq) + 2NO3(aq) + HgCl2(S) What is the Cl concentration in a 0.25-mL sample of normal serum that requires 1.46 mL of .000825 M Hg(NO3)2(aq) to reach the end point?

Answer 1

16.

Moles of Lithium = ( mass /molar mass) = (1.5/7) = 0.214

Moles of nitrogen = ( mass/molar mass) = (1.5/28) = 0.053

From the balanced reaction

6 Li + N₂$\rightarrow$ 2 Li₃N

Mole ratio of Li and N₂ = 6:1

So, 0.053 mole Nitrogen completely reacts with (0.053×6) = 0.318.

But moles of Li taken = 0.214.

As moles of Li is less than required , limiting reactant is Lithium.

17.

From the balanced reaction

Moles of AgNO₃ = moles of NaCl

Hence,

molar concentration of AgNO₃× volume of AgNO₃ solution

= molar concentration of NaCl × volume of NaCl solution

Or,

0.2503 × 20.22 = molar concentration of NaCl (C) × 15.00

Or, C = {(0.2503×20.22)/15} = 0.337 M

Hence, concentration of NaCl solution is 0.337 M.

18.

From the balanced reaction.

At end point

mmoles of Cl^- = 2 × mmoles of Hg(NO₃)₂.

Now, given

mmoles of Hg(NO₃)₂ = molarity × volume (mL)

= 0.000825 × 1.46 = 0.0012045

So, mmoles of Cl^- = (0.0012045 × 2 )= 0.002409

So , concentration of Cl^- in sample

= (mmoles of Cl^-/volume of Solution)

= (0.002409/0.25)

= 0.009636 M.