Question

1. $ce{na3(po4) + k(oh) -> na(oh) + k3(po4)}$

before chem rxn after chem rxn
$ce{_{\\_\\_\\_}na3(po4) + _{\\_\\_\\_}koh -> _{\\_\\_\\_}naoh + _{\\_\\_\\_}k3(po4)}$

Explanation:

Step1: Balance Na atoms

Count Na: 3 on left, 1 on right. Multiply NaOH by 3:
$\text{Na}_3(\text{PO}_4) + \text{KOH}
ightarrow 3\text{NaOH} + \text{K}_3(\text{PO}_4)$

Step2: Balance K atoms

Count K: 1 on left, 3 on right. Multiply KOH by 3:
$\text{Na}_3(\text{PO}_4) + 3\text{KOH}
ightarrow 3\text{NaOH} + \text{K}_3(\text{PO}_4)$

Step3: Verify all atoms

Check PO₄: 1 on each side. Check OH: 3 on each side. All atoms balanced.

Answer:

$\boldsymbol{1}\text{Na}_3(\text{PO}_4) + \boldsymbol{3}\text{KOH}
ightarrow \boldsymbol{3}\text{NaOH} + \boldsymbol{1}\text{K}_3(\text{PO}_4)$