Chapter 2 - Atoms and the Atomic Theory - Exercises - Law of Conservation of Mass - Page 60: 5

$m_{Ca(OH)_{2}} + m_{NH_{4}Cl} + m_{H_{2}O (l)} = m_{CaCl_{2}} + m_{NH_{3}(aq)} + m_{H_{2}O(aq)} + m_{H_{2}O (l)} $ 10.500 +11.125 + 62.316 = 14.336 + 69.605 83.941 = 83.941 So the Law of Conservation of Mass is obeyed.

$Ca(OH)_{2} (s) + 2 NH_{4}Cl (s) = CaCl_{2} (s)+ 2 NH_{3} (g)+ 2H_{2}O (g)$ According to the Law of Conservation of Mass $m_{reactants} = m_{products}$ $m_{Ca(OH)_{2}} + m_{NH_{4}Cl} = m_{CaCl_{2}} + m_{NH_{3}} + m_{H_{2}O}$ $m_{Ca(OH)_{2}}$ = 10.500 g $m_{NH_{4}Cl}$ = 11.125 g $m_{CaCl_{2}}$ = 14.336 g $10.500 + 11.125 = 14.336+ m_{NH_{3}} + m_{H_{2}O}$ $21.625 = 14.336+ m_{NH_{3}} + m_{H_{2}O}$ $m_{NH_{3}} + m_{H_{2}O} = 21.625 - 14.336$ $m_{NH_{3}} + m_{H_{2}O} = 7.289 g$......................................................................(1) $2 NH_{3} (g)+ 2H_{2}O (g) +H_{2}O (l) = 2 NH_{3} (aq)+ 2H_{2}O (aq) +H_{2}O (l)$ According to the Law of Conservation of Mass $m_{reactants} = m_{products}$ $ m_{NH_{3} (g)} + m_{H_{2}O (g)} + m_{H_{2}O (l)} = m_{NH_{3} (aq)} + m_{H_{2}O (aq)} + m_{H_{2}O (l)}$ $m_{H_{2}O (l)} $ = 62.316 g $m_{NH_{3} (aq)} + m_{H_{2}O (aq)} + m_{H_{2}O (l)}$ = 69.605 g $ m_{NH_{3} (g)} + m_{H_{2}O (g)} + 62.316 = 69.605$ $ m_{NH_{3} (g)} + m_{H_{2}O (g)} $= 69.605 - 62.316 $ m_{NH_{3} (g)} + m_{H_{2}O (g)} $= 7.289 g.................................................................(2) $ m_{NH_{3} (g)} + m_{H_{2}O (g)} $ at (1) and $ m_{NH_{3} (g)} + m_{H_{2}O (g)} $ at (2) are both 7.289 g So these observations conform the law of conservation of mass OR $Ca(OH)_{2} (s) + 2 NH_{4}Cl (s) = CaCl_{2} (s)+ 2 NH_{3} (g)+ 2H_{2}O (g)$ $2 NH_{3} (g)+ 2H_{2}O (g) +H_{2}O (l) = 2 NH_{3} (aq)+ 2H_{2}O (aq) +H_{2}O (l)$ Both reactions added gives: $Ca(OH)_{2} (s) + 2 NH_{4}Cl (s) + H_{2}O (l) = CaCl_{2} (s)+ 2 NH_{3} (aq)+ 2H_{2}O (aq) + H_{2}O (l)$ According to the Law of Conservation of Mass $m_{reactants} = m_{products}$ $m_{Ca(OH)_{2}} + m_{NH_{4}Cl} + m_{H_{2}O (l)} = m_{CaCl_{2}} + + m_{NH_{3}(aq)} + m_{H_{2}O(aq)} + m_{H_{2}O (l)} $ $m_{Ca(OH)_{2}}$ = 10.500 g $m_{NH_{4}Cl}$ = 11.125 g $m_{H_{2}O (l)} $ = 62.316 g $m_{CaCl_{2}}$ = 14.336 g $m_{NH_{3} (aq)} + m_{H_{2}O (aq)} + m_{H_{2}O (l)}$ = 69.605 g 10.500 +11.125 + 62.316 = 14.336 + 69.605 83.941 = 83.941 So the Law of Conservation of Mass is obeyed.