What is the mass of potassium chloride produc | vedclass

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Question

(C) The balanced chemical equation for the decomposition of potassium chlorate is: $2KClO_3 \longrightarrow 2KCl + 3O_2$ Calculate the molar mass of $

Vedclass · Accepted Answer

$7.45$

Vedclass · Answer

(C) The balanced chemical equation for the decomposition of potassium chlorate is: $2KClO_3 \longrightarrow 2KCl + 3O_2$ Calculate the molar mass of $KClO_3$: $M(KClO_3) = 39 + 35.5 + 3 	imes 16 = 122.5 \ g/mol$ Calculate the molar mass of $KCl$: $M(KCl) = 39 + 35.5 = 74.5 \ g/mol$ Calculate the moles of $KClO_3$ present: $n(KClO_3) = \frac{12.25 \ g}{122.5 \ g/mol} = 0.1 \ mol$ From the stoichiometry of the reaction,$2 \ mol$ of $KClO_3$ produces $2 \ mol$ of $KCl$. Therefore,$0.1 \ mol$ of $KClO_3$ will produce $0.1 \ mol$ of $KCl$. Calculate the mass of $KCl$ produced: $Mass = n 	imes M = 0.1 \ mol 	imes 74.5 \ g/mol = 7.45 \ g$

List-$I$	List-$II$ (At $STP$)
$(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$	$(i)$ $0.224 \ L \ CO_2$
$(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl} \text{reaction}$	$(ii)$ $4.48 \ L \ CO_2$
$(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iii)$ $0.448 \ L \ CO_2$
$(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iv)$ $2.24 \ L \ CO_2$
	$(v)$ $22.4 \ L \ CO_2$

What is the mass of potassium chloride produced when $12.25 \ g$ of potassium chlorate undergoes decomposition (in $g$)? (Atomic masses: $K=39, Cl=35.5, O=16$)

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