How many grams of KClO_3 must be decomposed t | vedclass

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(A) The decomposition reaction is: $2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$.First,calculate the moles of $O_2$ using the ideal gas law $PV = nRT$:$P

Vedclass · Accepted Answer

$7.8$

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(A) The decomposition reaction is: $2KClO_3(s) ightarrow 2KCl(s) + 3O_2(g)$.First,calculate the moles of $O_2$ using the ideal gas law $PV = nRT$:$P = 740 \, mm \, Hg = 740/760 \, atm \approx 0.9737 \, atm$.$V = 2.40 \, L$.$T = 25 + 273.15 = 298.15 \, K$.$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.$n(O_2) = (PV)/(RT) = (0.9737 	imes 2.40) / (0.0821 	imes 298.15) \approx 0.0954 \, mol$.From the stoichiometry,$2 \, mol$ of $KClO_3$ produces $3 \, mol$ of $O_2$.So,$n(KClO_3) = (2/3) 	imes n(O_2) = (2/3) 	imes 0.0954 \approx 0.0636 \, mol$.Mass of $KClO_3 = n 	imes 	ext{Molar mass} = 0.0636 	imes 122.5 \approx 7.79 \, g \approx 7.8 \, g$.

How many grams of $KClO_3$ must be decomposed to obtain $2.40 \, L$ of $O_2$ gas at $25 \, ^\circ C$ and $740 \, mm \, Hg$ pressure (in $, g$)? (Molar mass of $KClO_3 = 122.5 \, g \, mol^{-1}$)

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