fill in the blanks to balance this chemical reaction.
____ h3po4 + ____ ca(oh)2 → ____ ca3(po4)2 + ____ h2o

Question

Accepted Answer

# Explanation:
## Step1: Balance phosphorus atoms
In $H_3PO_4$ and $Ca_3(PO_4)_2$, to balance $P$ atoms, since there are $2$ $P$ atoms in $Ca_3(PO_4)_2$, we put a coefficient of $2$ in front of $H_3PO_4$.
$$2H_3PO_4+Ca(OH)_2
ightarrow Ca_3(PO_4)_2 + H_2O$$
## Step2: Balance calcium atoms
In $Ca(OH)_2$ and $Ca_3(PO_4)_2$, since there are $3$ $Ca$ atoms in $Ca_3(PO_4)_2$, we put a coefficient of $3$ in front of $Ca(OH)_2$.
$$2H_3PO_4 + 3Ca(OH)_2
ightarrow Ca_3(PO_4)_2+H_2O$$
## Step3: Balance hydrogen and oxygen atoms
Count $H$ and $O$ atoms. On the left - hand side, from $2H_3PO_4$ we have $2\times3 = 6$ $H$ atoms from $H_3PO_4$ and $3\times2=6$ $H$ atoms from $Ca(OH)_2$, so total $H$ atoms $=6 + 6=12$. From $2H_3PO_4$ we have $2\times4 = 8$ $O$ atoms (excluding $OH$ groups) and from $3Ca(OH)_2$ we have $3\times2 = 6$ $O$ atoms in $OH$ groups.
On the right - hand side, in $Ca_3(PO_4)_2$ there are $2\times4 = 8$ $O$ atoms (excluding $H_2O$). Let the coefficient of $H_2O$ be $x$.
For $H$ atoms: $x\times2=12$, so $x = 6$.
Check $O$ atoms: Left - hand side total $O$ atoms (excluding $H_2O$ calculation part) $=8 + 6=14$. Right - hand side, in $Ca_3(PO_4)_2$ $O$ atoms (excluding $H_2O$) $=8$, and in $6H_2O$ we have $6$ $O$ atoms. $8 + 6=14$

# Answer:
$2$ $H_3PO_4+3$ $Ca(OH)_2
ightarrow1$ $Ca_3(PO_4)_2 + 6$ $H_2O$

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QUESTION IMAGE

Question

Explanation:

Step1: Balance phosphorus atoms

Step2: Balance calcium atoms

Step3: Balance hydrogen and oxygen atoms

Answer: