Question

lars is balancing equations with his study group. he is unsure about one equation because each member of the study group came up with a different answer. which is the proper way to balance the equation ca(oh)₂ + h₃po₄ → ca₃(po₄)₂ + h₂o?
o ca(oh)₂ + h₃po₄ → ca₃(po₄)₂ + 6h₂o
o 3ca(oh)₂ + 2h₃po₄ → ca₃(po₄)₂ + 6h₂o
o 3ca(oh)₂ + 2h₃po₄ → ca₃(po₄)₂ + 3h₂o
o ca(oh)₂ + 2h₃po₄ → ca₃(po₄)₂ + 6h₂o

Explanation:

Step1: Balance calcium atoms

On the right - hand side, there are 3 calcium atoms in $Ca_3(PO_4)_2$. So, we put a coefficient of 3 in front of $Ca(OH)_2$ on the left - hand side: $3Ca(OH)_2+H_3PO_4
ightarrow Ca_3(PO_4)_2 + H_2O$.

Step2: Balance phosphate groups

There are 2 phosphate groups in $Ca_3(PO_4)_2$. So, we put a coefficient of 2 in front of $H_3PO_4$ on the left - hand side: $3Ca(OH)_2 + 2H_3PO_4
ightarrow Ca_3(PO_4)_2+H_2O$.

Step3: Balance hydrogen and oxygen atoms

On the left - hand side, from $3Ca(OH)_2$ we have 6 oxygen atoms from the hydroxide groups and from $2H_3PO_4$ we have 8 oxygen atoms from the phosphate groups, a total of 14 oxygen atoms in the reactants. On the right - hand side, in $Ca_3(PO_4)_2$ there are 8 oxygen atoms in the phosphate groups. So, in $H_2O$ we need 6 oxygen atoms, which means we put a coefficient of 6 in front of $H_2O$. Also, for hydrogen, on the left - hand side we have $3\times2$ (from $Ca(OH)_2$) + $2\times3$ (from $H_3PO_4$) = 12 hydrogen atoms, and on the right - hand side $6\times2$ (from $H_2O$) = 12 hydrogen atoms. The balanced equation is $3Ca(OH)_2+2H_3PO_4
ightarrow Ca_3(PO_4)_2 + 6H_2O$.

Answer:

B. $3Ca(OH)_2 + 2H_3PO_4
ightarrow Ca_3(PO_4)_2+6H_2O$