A mixture of 100 mmol of Ca(OH)_2 and 2 g o | vedclass

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(A) The reaction is: $Ca(OH)_2 + Na_2SO_4 	o CaSO_4(s) + 2NaOH$.Initial moles of $Ca(OH)_2 = 100 \ mmol = 0.1 \ mol$.Initial moles of $Na_2SO_4 = \fr

Vedclass · Accepted Answer

$1.9 \ g, 0.28 \ mol \ L^{-1}$

Vedclass · Answer

(A) The reaction is: $Ca(OH)_2 + Na_2SO_4 	o CaSO_4(s) + 2NaOH$.Initial moles of $Ca(OH)_2 = 100 \ mmol = 0.1 \ mol$.Initial moles of $Na_2SO_4 = \frac{2 \ g}{142 \ g \ mol^{-1}} \approx 0.014 \ mol = 14 \ mmol$.Since $Na_2SO_4$ is the limiting reagent,$14 \ mmol$ of $CaSO_4$ will be formed.Mass of $CaSO_4 = 14 	imes 10^{-3} \ mol 	imes 136 \ g \ mol^{-1} = 1.904 \ g \approx 1.9 \ g$.In the reaction,$2 \ mmol$ of $NaOH$ are produced for every $1 \ mmol$ of $Na_2SO_4$ consumed.Moles of $OH^{-} = 2 	imes 14 \ mmol = 28 \ mmol = 0.028 \ mol$.Concentration of $OH^{-} = \frac{0.028 \ mol}{0.1 \ L} = 0.28 \ mol \ L^{-1}$.

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