Answer:
Coefficient of [tex]H_{2}O[/tex] in the balanced equation with smallest possible integer is 6.
Explanation:
Unbalanced equation: [tex]Ca(OH)_{2}(aq.)+H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+H_{2}O(l)[/tex]
Balance Ca: [tex]3Ca(OH)_{2}(aq.)+H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+H_{2}O(l)[/tex]
Balance [tex]PO_{4}[/tex] : [tex]3Ca(OH)_{2}(aq.)+2H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+H_{2}O(l)[/tex]
Balance H and O: [tex]3Ca(OH)_{2}(aq.)+2H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+6H_{2}O(l)[/tex]
Balanced equation: [tex]3Ca(OH)_{2}(aq.)+2H_{3}PO_{4}(aq.)\rightarrow Ca_{3}(PO_{4})_{2}(s)+6H_{2}O(l)[/tex]
So coefficient of [tex]H_{2}O[/tex] in the balanced equation with smallest possible integer is 6.
