How many milli-Iitre of 0.5 N SnCl2 solution will reduce 600 mL of 0.1 N HgCl2 to Hg2Cl2?

SnCl2+2+Hg+2Cl2⟶Hg2+1Cl2+Sn +40.5 x V(mL) = 600 x0.1V(mL) =120 mL

How many milli-Iitre of 0.5 N SnCl2 solution will reduce 600 mL of 0.1 N HgCl2 to Hg2Cl2?

SnCl2+2+Hg+2Cl2⟶Hg2+1Cl2+Sn +40.5 x V(mL) = 600 x0.1V(mL) =120 mL

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How many milli-Iitre of 0.5 N SnCl₂ solution will reduce 600 mL of 0.1 N HgCl₂ to Hg₂Cl₂?

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120 mL

240 mL

60 mL

30 mL

answer is A.

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Detailed Solution

$\overset{+ 2}{Sn {Cl}_{2}} + \overset{+ 2}{Hg} {Cl}_{2} ⟶ \overset{+ 1}{{Hg}_{2}} {Cl}_{2} + \overset{^{+ 4}}{Sn}$
0.5 x V(mL) = 600 x0.1
V(mL) =120 mL

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How many milli-Iitre of 0.5 N SnCl2 solution will reduce 600 mL of 0.1 N HgCl2 to Hg2Cl2?

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How many milli-Iitre of 0.5 N SnCl₂ solution will reduce 600 mL of 0.1 N HgCl₂ to Hg₂Cl₂?