What is the weight in g of 90% pure H_2SO_4 r | vedclass

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Question

(D) The balanced chemical equation for the reaction is:$2NaOH + H_2SO_4 	o Na_2SO_4 + 2H_2O$From the equation,$2 \ mol$ of $NaOH$ $(2 	imes 40 = 80

Vedclass · Accepted Answer

$6.81$

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(D) The balanced chemical equation for the reaction is:$2NaOH + H_2SO_4 	o Na_2SO_4 + 2H_2O$From the equation,$2 \ mol$ of $NaOH$ $(2 	imes 40 = 80 \ g)$ reacts with $1 \ mol$ of $H_2SO_4$ $(98 \ g)$.Moles of $NaOH$ in $5 \ g = \frac{5}{40} = 0.125 \ mol$.According to the stoichiometry,$2 \ mol$ of $NaOH$ requires $1 \ mol$ of $H_2SO_4$.Therefore,$0.125 \ mol$ of $NaOH$ requires $\frac{1}{2} 	imes 0.125 = 0.0625 \ mol$ of $H_2SO_4$.Mass of pure $H_2SO_4 = 0.0625 \ mol 	imes 98 \ g/mol = 6.125 \ g$.Since the $H_2SO_4$ is $90\%$ pure,the required weight is $\frac{6.125}{0.90} \approx 6.81 \ g$.

What is the weight in $g$ of $90\%$ pure $H_2SO_4$ required to neutralize $5 \ g$ of caustic soda?

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