The ammonia prepared by treating ammonium sul | vedclass

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(B) The reaction for the preparation of ammonia is: $(NH_4)_2SO_4 + Ca(OH)_2 \longrightarrow CaSO_4 \cdot 2H_2O + 2NH_3$. For $1584 \ g$ of $(NH_4)_2S

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$2992$

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(B) The reaction for the preparation of ammonia is: $(NH_4)_2SO_4 + Ca(OH)_2 \longrightarrow CaSO_4 \cdot 2H_2O + 2NH_3$. For $1584 \ g$ of $(NH_4)_2SO_4$ $(M = 132 \ g \ mol^{-1})$,the number of moles $n = 1584 / 132 = 12 \ mol$. This produces $12 \ mol$ of gypsum $(CaSO_4 \cdot 2H_2O)$ and $24 \ mol$ of $NH_3$. Mass of gypsum $(M = 172 \ g \ mol^{-1})$ $= 12 	imes 172 = 2064 \ g$. The reaction with nickel chloride is: $NiCl_2 \cdot 6H_2O + 6NH_3 \longrightarrow [Ni(NH_3)_6]Cl_2 + 6H_2O$. For $952 \ g$ of $NiCl_2 \cdot 6H_2O$ $(M = 238 \ g \ mol^{-1})$,the number of moles $n = 952 / 238 = 4 \ mol$. This reacts with $24 \ mol$ of $NH_3$ to produce $4 \ mol$ of $[Ni(NH_3)_6]Cl_2$. Mass of $[Ni(NH_3)_6]Cl_2$ $(M = 232 \ g \ mol^{-1})$ $= 4 	imes 232 = 928 \ g$. Total mass $= 2064 \ g + 928 \ g = 2992 \ g$.

Similar Questions

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