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?\nca(oh)2 + h3po4 → ca3(po4)2 + 6h2o\n3ca(oh)2 + 2h3po4 → ca3(po4)2 + 6h2o\n3ca(oh)2 + 2h3po4 → ca3(po4)2 + 3h2o\nca(oh)2 + 2h3po4 → ca3(po4)2 + 6h2o
Answer
Explanation:
Step1: Balance calcium atoms
On the right - hand side of the equation $Ca_3(PO_4)_2$, there are 3 calcium atoms. So, we need 3 moles of $Ca(OH)_2$ on the left - hand side to balance the calcium atoms. The equation becomes $3Ca(OH)_2+H_3PO_4\rightarrow Ca_3(PO_4)_2 + H_2O$.
Step2: Balance phosphate atoms
On the right - hand side, there are 2 phosphate ($PO_4$) groups in $Ca_3(PO_4)_2$. So, we need 2 moles of $H_3PO_4$ on the left - hand side to balance the phosphate atoms. The equation is now $3Ca(OH)_2 + 2H_3PO_4\rightarrow Ca_3(PO_4)_2+H_2O$.
Step3: Balance hydrogen and oxygen atoms
On the left - hand side, in $3Ca(OH)_2$ there are 6 moles of $OH^-$ groups and in $2H_3PO_4$ there are 6 moles of $H^+$ atoms. Combining them gives 12 moles of hydrogen atoms and 6 moles of oxygen atoms from the $OH^-$ groups. On the right - hand side, in $Ca_3(PO_4)_2$ there are no hydrogen or oxygen atoms from the $Ca$ and $PO_4$ part. So, we get 6 moles of $H_2O$ to balance the hydrogen and oxygen atoms. 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 in the multiple - choice list)