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Stoichiometric Calculations

The ammonia p...

Question

The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by $N i C l_{2} \cdot 6 H_{2} O$ to form a stable coordination compound. Assume that both the reactions are $100 %$ complete. If $1584$ g of ammonium sulphate and $952$ g of $N i C l_{2} \cdot 6 H_{2} O$ are used in the preparation, the combined weight(in grams) of gypsum and the nickel-ammonia coordination compound thus produced is _________.
(Atomic weights in $g m o l^{- 1} : H = 1, N = 14, O = 16, S = 32, C l = 35.5, C a = 40, N i = 59$ )

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Solution

The answer is $2992$ .

$(N H_{4})_{2} S O_{4} 1584 g = 12 m o l + C a (O H)_{2} \to C a S O_{4} \cdot 2 H_{2} O g y p s u m (M = 172) 12 m o l + 2 N H_{3} 24 m o l e$

$N i C l_{2} \cdot 6 H_{2} O 952 g = 4 m o l + 6 N H_{3} 24 m o l \to [N i (N H_{3})_{6}] (M = 232) 4 m o l C l_{2} + 6 H_{2} O$

$T o t a l m a s s = 12 \times 172 + 4 \times 232 = 2992 g$ .

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Similar questions

N i C l_{2} .6 H_{2} O

[N i (N H_{3})_{6}] C l_{2}

100

1584

952

N i C l_{2} .6 H_{2} O

C a S O_{4} .2 H_{2} O

C a (O H)_{2} + (N H_{4})_{2} S O_{4} \to 2 N H_{3} + C a S O_{4} .2 H_{2} O

N i C l_{2} .6 H_{2} O + 6 N H_{3} \to [N i (N H_{3})_{6}] C l_{2} + 6 H_{2} O

g m o l^{- 1}

H = 1, N = 14, O = 16, S = 32, C l = 35.5, C a = 40, N i = 59)

N i C l_{2} .6 H_{2} O

[N i (N H_{3})_{6}] C l_{2}

100

1584

952

N i C l_{2} .6 H_{2} O

C a S O_{4} .2 H_{2} O

C a (O H)_{2} + (N H_{4})_{2} S O_{4} \to 2 N H_{3} + C a S O_{4} .2 H_{2} O

N i C l_{2} .6 H_{2} O + 6 N H_{3} \to [N i (N H_{3})_{6}] C l_{2} + 6 H_{2} O

g m o l^{- 1}

H = 1, N = 14, O = 16, S = 32, C l = 35.5, C a = 40, N i = 59)

59 g

[N = 14, H = 1, S = 32, O = 16]

[2 N H_{4} C l + C a (O H)_{2} \to C a C l_{2} + 2 H_{2} O + 2 N H_{3}]

The percentage of nitrogen by mass in ammonium sulphate is closest to (atomic masses H = 1, N = 14, O = 16, S = 32)

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