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  •  Consider the reaction <img alt="\mathrm{4 \mathrm{HNO}_{3}(\mathrm{l})+3 \mathrm{KCl}(\mathrm{s}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{NOCl}(\mathrm{g})+2 \mathrm{H}_{2} \mat

 Consider the reaction \mathrm{4 \mathrm{HNO}_{3}(\mathrm{l})+3 \mathrm{KCl}(\mathrm{s}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{NOCl}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{KNO}_{3}(\mathrm{~s})} The amount of  \mathrm{\mathrm{HNO}_{3} }  required to produce \mathrm{110.0 \mathrm{~g} }  of  \mathrm{\mathrm{KNO}_{3} }  is \mathrm{\text { (Given: Atomic masses of } \mathrm{H}, \mathrm{O}, \mathrm{N} \text { and } \mathrm{K} \text { are } 1,16,14 \text { and } 39 \text {, respectively.) } }

Option: 1

\mathrm{32.2 \mathrm{~g}}


Option: 2

\mathrm{69.4 \mathrm{~g}}


Option: 3

\mathrm{91.5 \mathrm{~g}}


Option: 4

\mathrm{162.5 \mathrm{~g}}


Answers (1)

best_answer

Given Reaction is 

\mathrm{\underset{xg}{4 \mathrm{HNO}_{3}(\mathrm{l})}+3 \mathrm{KCl}(\mathrm{s}) \longrightarrow \mathrm{Cl}_{2}(\mathrm{g})+\mathrm{NOCl}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\underset{110g}{3 \mathrm{KNO}_{3}(\mathrm{s})}}

According to Stoichiometry

 \mathrm{\frac{m_{HNO_{3}}}{4}= \frac{m_{KNO_{3}}}{3}}


\mathrm{m_{HNO_{3}}= 63g/mol \quad m_{KNO_{3}}= 101 g/mol}


\mathrm{\frac{w_{HNO_{3}}}{63\times 4}= \frac{110g}{101\times 3}}

\\ \mathrm{\frac{w_{HNO_{3}}}{63\times 4}= \frac{110g}{101\times 3}} \\ \\ \mathrm{w_{HNO_{3}}= 91.48 g}
\mathrm{w_{HNO_3}= 91.5 g}
The correct answer is option (3).

Posted by

Divya Prakash Singh

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