Sulphuric acid reacts with sodium hydroxide a | vedclass

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(C) The balanced chemical equation is: $H_2SO_4 + 2NaOH \longrightarrow Na_2SO_4 + 2H_2O$Initial moles of $H_2SO_4 = 	ext{Molarity} 	imes 	ext{Volu

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$14.2$

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(C) The balanced chemical equation is: $H_2SO_4 + 2NaOH \longrightarrow Na_2SO_4 + 2H_2O$Initial moles of $H_2SO_4 = 	ext{Molarity} 	imes 	ext{Volume} = 0.2 \ M 	imes 1 \ L = 0.2 \ mol$.Initial moles of $NaOH = 	ext{Molarity} 	imes 	ext{Volume} = 0.2 \ M 	imes 1 \ L = 0.2 \ mol$.According to the stoichiometry, $1 \ mol$ of $H_2SO_4$ requires $2 \ mol$ of $NaOH$.Therefore, $0.2 \ mol$ of $H_2SO_4$ would require $0.4 \ mol$ of $NaOH$.Since only $0.2 \ mol$ of $NaOH$ is available, $NaOH$ is the limiting reagent.From the stoichiometry, $2 \ mol$ of $NaOH$ produces $1 \ mol$ of $Na_2SO_4$.So, $0.2 \ mol$ of $NaOH$ will produce $0.1 \ mol$ of $Na_2SO_4$.Molar mass of $Na_2SO_4 = (2 	imes 23) + 32 + (4 	imes 16) = 142 \ g/mol$.Mass of $Na_2SO_4 = 	ext{moles} 	imes 	ext{molar mass} = 0.1 \ mol 	imes 142 \ g/mol = 14.2 \ g$.

Sulphuric acid reacts with sodium hydroxide as follows:
$H_2SO_4 + 2NaOH \longrightarrow Na_2SO_4 + 2H_2O$
What will be the amount of sodium sulphate formed, when $1 \ L$ of $0.2 \ M$ sulphuric acid is allowed to react with $1 \ L$ of $0.2 \ M$ sodium hydroxide solution (in $g$)?

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Sulphuric acid reacts with sodium hydroxide as follows:$H_2SO_4 + 2NaOH \longrightarrow Na_2SO_4 + 2H_2O$What will be the amount of sodium sulphate formed, when $1 \ L$ of $0.2 \ M$ sulphuric acid is allowed to react with $1 \ L$ of $0.2 \ M$ sodium hydroxide solution (in $g$)?