One litre of 0.15 M Na_2SO_3 aqueous solutio | vedclass

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(D) The balanced chemical equation is:$3Na_2SO_3 + K_2Cr_2O_7 + 4H_2SO_4 \rightarrow 3Na_2SO_4 + K_2SO_4 + Cr_2(SO_4)_3 + 4H_2O$Moles of $Na_2SO_3 = 0

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$0.05$

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(D) The balanced chemical equation is:$3Na_2SO_3 + K_2Cr_2O_7 + 4H_2SO_4 ightarrow 3Na_2SO_4 + K_2SO_4 + Cr_2(SO_4)_3 + 4H_2O$Moles of $Na_2SO_3 = 0.15 \ M 	imes 1 \ L = 0.15 \ mol$Moles of $K_2Cr_2O_7 = 0.2 \ M 	imes 0.5 \ L = 0.1 \ mol$From the stoichiometry,$3 \ mol$ of $Na_2SO_3$ react with $1 \ mol$ of $K_2Cr_2O_7$.Moles of $K_2Cr_2O_7$ required for $0.15 \ mol$ of $Na_2SO_3 = \frac{0.15}{3} = 0.05 \ mol$.Moles of $K_2Cr_2O_7$ remaining $= 0.1 \ mol - 0.05 \ mol = 0.05 \ mol$.

One litre of $0.15 \ M$ $Na_2SO_3$ aqueous solution is mixed with $500 \ mL$ of $0.2 \ M$ $K_2Cr_2O_7$ aqueous solution in acid medium. What is the number of moles of $K_2Cr_2O_7$ remaining in the solution after the reaction?

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