For the following reactions: I. (NH4)2SO4+2NaOH→40% Na2SO3+2H2O+2NH3II. NH3+HCl→80%NH4ClIf 4g of NaOH is taken then

nNa2SO4=12×40100×440=2100=2×10{2}, nNH3=12×40100×440×2=4100×440, nNH4Cl=40100×440×80100=32×10−3, 32×10−32×10−2=1.6 times⇒ nNaOH>nHCl.

For the following reactions: I. (NH4)2SO4+2NaOH→40% Na2SO3+2H2O+2NH3II. NH3+HCl→80%NH4ClIf 4g of NaOH is taken then

nNa2SO4=12×40100×440=2100=2×10{2}, nNH3=12×40100×440×2=4100×440, nNH4Cl=40100×440×80100=32×10−3, 32×10−32×10−2=1.6 times⇒ nNaOH>nHCl.

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For the following reactions:
I. ${(N H_{4})}_{2} S O_{4} + 2 N a O H \overset{40 %}{\to} N a_{2} S O_{3} + 2 H_{2} O + 2 N H_{3}$
II. $N H_{3} + H C l \overset{80 %}{\to} N H_{4} C l$
If $4 g$ of $N a O H$ is taken then

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Produced moles of $N H_{4} C l$ (in reaction $I I^{n d}$ reaction) are 1.6 times of produced moles of $N a_{2} S O_{4}$ (in reaction I).

Reacting moles of $H C l$ (in reaction II) is lesser than reacting moles $N a O H$ (in reaction I).

Reacting moles of $H C l$ (in reaction II) is 20% lesser than original ${(N H_{4})}_{2} S O_{4}$ moles.

Produced mass of $N H_{4} C l$ is $2.71 g .$

answer is A, C.

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Detailed Solution

$n N a_{2} S O_{4} = \frac{1}{2} \times \frac{40}{100} \times \frac{4}{40} = \frac{2}{100} = 2 \times 10^{{2}}, n N H_{3} = \frac{1}{2} \times \frac{40}{100} \times \frac{4}{40} \times 2 = \frac{4}{100} \times \frac{4}{40}, n N H_{4} C l = \frac{40}{100} \times \frac{4}{40} \times \frac{80}{100} = 32 \times 10^{- 3}, \frac{32 \times 10^{- 3}}{2 \times 10^{- 2}} = 1.6 t i m e s \Rightarrow n_{N a O H} > n_{H C l} .$