lars is balancing equations with his study group. he is unsure about one equation because each member of the…

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?\no ca(oh)₂ + h₃po₄ → ca₃(po₄)₂ + 6h₂o\no 3ca(oh)₂ + 2h₃po₄ → ca₃(po₄)₂ + 6h₂o\no 3ca(oh)₂ + 2h₃po₄ → ca₃(po₄)₂ + 3h₂o\no ca(oh)₂ + 2h₃po₄ → ca₃(po₄)₂ + 6h₂o

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?\no ca(oh)₂ + h₃po₄ → ca₃(po₄)₂ + 6h₂o\no 3ca(oh)₂ + 2h₃po₄ → ca₃(po₄)₂ + 6h₂o\no 3ca(oh)₂ + 2h₃po₄ → ca₃(po₄)₂ + 3h₂o\no ca(oh)₂ + 2h₃po₄ → ca₃(po₄)₂ + 6h₂o

Answer

Explanation:

Step1: Balance calcium atoms

We have 1 Ca in $Ca(OH)_2$ and 3 Ca in $Ca_3(PO_4)_2$. So we put a coefficient of 3 in front of $Ca(OH)_2$: $3Ca(OH)_2 + H_3PO_4\rightarrow Ca_3(PO_4)_2+H_2O$.

Step2: Balance phosphate groups

We have 1 $PO_4$ in $H_3PO_4$ and 2 $PO_4$ in $Ca_3(PO_4)_2$. So we put a coefficient of 2 in front of $H_3PO_4$: $3Ca(OH)_2 + 2H_3PO_4\rightarrow Ca_3(PO_4)_2+H_2O$.

Step3: Balance hydrogen and oxygen atoms

On the left - hand side, we have $3\times2 = 6$ O from $Ca(OH)_2$ and $2\times4=8$ O from $2H_3PO_4$, a total of 14 O, and $3\times2 + 2\times3=12$ H. On the right - hand side, in $Ca_3(PO_4)_2$ there are 8 O. So in $H_2O$, we need 6 O and 12 H. So we put a coefficient of 6 in front of $H_2O$. The balanced equation is $3Ca(OH)_2 + 2H_3PO_4\rightarrow Ca_3(PO_4)_2+6H_2O$.

Answer:

$3Ca(OH)_2 + 2H_3PO_4\rightarrow Ca_3(PO_4)_2+6H_2O$ (the second option)