Consider the reaction4 HNO_{3(ell)} + 3 KCl_{ | vedclass

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(C) The balanced chemical equation is:$4 HNO_{3} + 3 KCl \rightarrow Cl_{2} + NOCl + 2 H_{2}O + 3 KNO_{3}$Calculate the molar masses:$M(HNO_{3}) = 1 +

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$91.5$

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(C) The balanced chemical equation is:$4 HNO_{3} + 3 KCl ightarrow Cl_{2} + NOCl + 2 H_{2}O + 3 KNO_{3}$Calculate the molar masses:$M(HNO_{3}) = 1 + 14 + (3 	imes 16) = 63 \ g/mol$$M(KNO_{3}) = 39 + 14 + (3 	imes 16) = 101 \ g/mol$From the stoichiometry,$3 \ moles$ of $KNO_{3}$ are produced by $4 \ moles$ of $HNO_{3}$.Number of moles of $KNO_{3}$ produced = $\frac{110.0 \ g}{101 \ g/mol} \approx 1.089 \ mol$.Moles of $HNO_{3}$ required = $\frac{4}{3} 	imes (	ext{moles of } KNO_{3}) = \frac{4}{3} 	imes \frac{110}{101} \approx 1.452 \ mol$.Mass of $HNO_{3}$ required = $1.452 \ mol 	imes 63 \ g/mol = 91.5 \ g$.

Consider the reaction
$4 HNO_{3(\ell)} + 3 KCl_{(s)} \rightarrow Cl_{2(g)} + NOCl_{(g)} + 2 H_{2}O_{(g)} + 3 KNO_{3(s)}$
The amount of $HNO_{3}$ required to produce $110.0 \ g$ of $KNO_{3}$ is $...... \ g$.
(Given: Atomic masses of $H, O, N$ and $K$ are $1, 16, 14$ and $39$ respectively.)

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Consider the reaction$4 HNO_{3(\ell)} + 3 KCl_{(s)} \rightarrow Cl_{2(g)} + NOCl_{(g)} + 2 H_{2}O_{(g)} + 3 KNO_{3(s)}$The amount of $HNO_{3}$ required to produce $110.0 \ g$ of $KNO_{3}$ is $...... \ g$.(Given: Atomic masses of $H, O, N$ and $K$ are $1, 16, 14$ and $39$ respectively.)