Question

(1) X30 pnts) Mixing together NH(aq) and CIO (aq) liberate a number of toxic gases. The following reactions are the dominant pathways for the production of toxic gasses: 2010 (aq) + 2NH(aq) Cl(aq) + 2NH, (aq) + H2O(1) 2NH3(aq) + Cl:(aq) + 2NH2Cl(g) Cl(aq) Cl.(g) "The amount of NH Cl(o) produced depends on the amount of Cl that remains in solution. 275 mL of a 0.155 M aqueous solution of NaCIO is fully reacted with excess NHCl(aq) generating a mixture of NH,Cl and Cl2 gas. The gasses were directed into an evacuated 1.00 L ridged container at -25,0 °C. If the partial pressure of Cl gas is 0.314 atm, what is the partial pressure of NHCl(g) produced, and the total pressure in the container (assume all Cl that remains in solution reacts with ammonia)?

Answer 1

Let's first calculate the number of moles of NaClO initially added:

$n=M\cdot V(L)=0.155\frac{mol}{L}\cdot 0.275L=0.0426moles$

And using the ideal gas equation we can calculate the number of moles of Cl₂ generated:

$n=\frac{pV}{RT}=\frac{0.314atm\cdot 1.00L}{0.082\frac{L\cdot atm}{mol\cdot K}\cdot 248K}=0.0154moles$

If we take a look at the first reaction, we can see that the number of moles of Cl₂ generated is half the number of moles of NaClO added, so the total number of Cl₂ moles obtained was:

$n_{Cl_{2}}=\frac{0.0426moles}{2}=0.0213moles$

Of these, 0.0154 were evaporated as Cl₂ gas and the remaining 0.0059 moles reacted with ammonia (of which there are the same number of moles as initial moles of NaClO). Since each mole of Cl₂ that reacts with ammonia produces 2 moles of NH₂Cl, the number of moles of this product is double the number of moles of Cl₂ reacting; that is: 0.0118 moles.

We can now calculate the pressure generated by these moles:

$p=\frac{nRT}{V}=\frac{0.0118moles\cdot 0.082\frac{L\cdot atm}{mol\cdot K}\cdot 248K}{1.00L}=0.240atm$

The total pressure is the sum of partial pressures: 0.240 atm + 0.314 atm = 0.554 atm.