A 0.608-g sample of fertilizer contained nitrogen as ammonium sulfate, (NH4)2SO4. It was analyzed for nitrogen by
Question:
(NH4)2SO4(s) + 2NaOH(aq) → Na2SO4(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) → NH4Cl(aq)
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From the equations NH 3 HCl NH 4 Cl and NaOH HCl NaCl H 2 O we write Mol NH 3 mol ...View the full answer
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