Question

part d
bacl₂(aq) + h₃po₄(aq) → ba₃(po₄)₂(s) + hcl(aq)
express your answers as integers separated by commas.
6,4,2,12
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incorrect; try again
those are not the smallest whole - number coefficients that could be used to balance this equation. simplify your coefficients.

Explanation:

Step1: Balance barium atoms

To balance barium ($Ba$) atoms, since there are 3 $Ba$ atoms in $Ba_3(PO_4)_2$, we put a coefficient of 3 in front of $BaCl_2$: $3BaCl_2(aq)+H_3PO_4(aq)
ightarrow Ba_3(PO_4)_2(s)+HCl(aq)$

Step2: Balance phosphate groups

To balance the phosphate ($PO_4$) groups, since there are 2 $PO_4$ groups in $Ba_3(PO_4)_2$, we put a coefficient of 2 in front of $H_3PO_4$: $3BaCl_2(aq) + 2H_3PO_4(aq)
ightarrow Ba_3(PO_4)_2(s)+HCl(aq)$

Step3: Balance chlorine and hydrogen atoms

After balancing $Ba$ and $PO_4$, for the chlorine ($Cl$) and hydrogen ($H$) atoms, we put a coefficient of 6 in front of $HCl$. The balanced equation is $3BaCl_2(aq)+2H_3PO_4(aq)
ightarrow Ba_3(PO_4)_2(s)+6HCl(aq)$

Answer:

3,2,1,6