Can someone give me some directions as to how to reach theanswer for this proble

Question

Can someone give me some directions as to how to reach theanswer for this problem:

A 0.608g sample of fertilizer contained nitrogen as ammoniumsulfate,

(NH4)2 SO4(s)+2NaOH(aq)àNa2SO4(aq) + 2H2O(l)+2NH3(g)

The ammonia was collected in 46.3 mL of 0.213 M HCL(hydrochloric acid), with which it reacted.

NH3(g)+HCL(aq)àNH4Cl(aq)

This solution was titrated for excess hydrochloric acid with44.3 mL of 0.128 M NaOH

NaOH(aq)+HCL(aq)àNaCL(aq)+H2O(l)

What is the percentage of nitrogen in the fertilizer?

Well the answer was given that it is 9.66% but how do I go aboutsolving it so I can finish other similar problems like it?

Explanation / Answer

the only step were excess wasn't used was in thetitration. So You can use that to work your way back to theanswer. since teh reaction is one to one then to find the amount ofexcess HCl there was .0443L * .128 mol/L=5.67*10^-3 mol Then find how much HCl you used and subtract the excess to getthe amount of NH3 there was. .0463 L* .213 mol/L=9.86*10^-3-5.67*10^-3=4.19*10^-3 molNH3 Now you dont' need to use the first equation just usethis and the molar mass of NH3 to find the number of grams ofN in the sample. MM NH3=17.0307g/mol.   % N=82.25% 4.19*10^-3*17.030=.07135g * .8225= .05869 g N in sample .05869/.608 *100 =9.66%

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