1 mol BaF2 - 2 mol H2SO4 - resulting mixture- which will be neutralised by-

The correct option is B 2 mol Ca(OH)_2 BaF_2 (1 mol) +H_2SO_4 (2 mol) → BaSO_4 + H_2F_2 + H_2SO_4 (unreacted) 1 #160; #160; #160; # ...

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Stoichiometric Calculations

1 mol BaF2 ...

Question

$1$ mol $B a F_{2}$ $+ 2$ mol $H_{2} S O_{4}$ $\to$ resulting mixture, which will be neutralised by:

1

mol

K O H

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2

mol

C a (O H)_{2}

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2

mol

N a O H

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2

mol

K O H

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Solution

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Similar questions

H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H = - 68.39

K + H_{2} O \to K O H (a q) + \frac{1}{2} H_{2}; Δ H = - 48.0

K O H + H_{2} O \to K O H (a q); Δ H = - 14.0

K O H

0.2 M H_{2} {S O}_{4}

{N a O H (1 m o l) + C a (O H)}_{2} (2 m o l)

H_{3} P O_{2}, H_{3} P O_{3}

H_{3} P O_{4}

x

N a O H,' y' mol of Ca {(O H)}_{2}

' z' mol of Al {(O H)}_{3}

x, y

z

\begin{matrix} Column - I & Column - II (Enthalpy change in kcal) & (Neutralisation) (A) & < 13.7 kcal & (p) & H C l (1 m o l) + N a O H (1 m o l) (B) & = 13.7 kcal & (q) & H F (1 m o l) + N a O H (1 m o l) (C) & > 13.7 kcal & (r) & N H_{4} O H (1 m o l) + H C l (1 m o l) (D) & = 27.4 kcal & (s) & N a O H (2 m o l) + H_{2} S O_{4} (1 m o l) (t) & N a O H (1 m o l) + C H_{3} C O O H (1 m o l) \end{matrix}

^{2}

^{- 1}

^{2}

^{- 1}

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