White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of HCl obtained by the hydrolysis of the product formed by the reaction of 62 g of white phosphorus with chlorine in the presence of water

<p><span data-teams="true">This is a short answer type question as classified in NCERT Exemplar</span></p><p>P₄+6Cl₂→4PCl₃…&nbsp; (i)</p><p>PCl₃+3H₂O→H₃PO₃+3HCl ×4</p><p>On adding eq. (i) and (ii)</p><p>P₄+6Cl₂+12H₂O→4H₃PO₃+12HCl</p><p>1&nbsp;mol&nbsp;of white phosphorus produces&nbsp;12&nbsp;mol&nbsp;of&nbsp;HCl</p><p>62&nbsp;g&nbsp;of white phosphorus has been taken which is equivalent to<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi mathvariant="normal">N</mi> <mi mathvariant="normal">o</mi> <mo>.</mo> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">f</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">M</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">&nbsp;</mi> <mi mathvariant="normal">&nbsp;</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">V</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal">u</mi> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">&nbsp;</mi> </mrow> </mrow> </mfrac> </math> </span> = <span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi mathvariant="normal">N</mi> <mi mathvariant="normal">o</mi> <mo>.</mo> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">f</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">M</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">&nbsp;</mi> <mi mathvariant="normal">&nbsp;</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">M</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">y</mi> </mrow> </mrow> </mfrac> </math> </span>mol. Therefore&nbsp;6&nbsp;molHCl&nbsp;will be formed.</p><p>Mass of 6 molHCl=6×36.5=219.0g HCl</p>

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