Question

1. _ baf₂ + _ k₃po₃ → _ kf + _ ba₃(po₃)₃

Explanation:

Step1: Balance Ba atoms

Let the coefficient of \(BaF_2\) be \(x\) and of \(Ba_3(PO_3)_2\) be \(y\). For Ba balance: \(x = 3y\). Let \(y = 1\), then \(x=3\).

Step2: Balance F atoms

Since coefficient of \(BaF_2\) is \(3\) (from Step 1), number of F atoms from \(BaF_2\) is \(3\times2 = 6\). Let the coefficient of \(KF\) be \(z\). For F balance: \(z = 6\).

Step3: Balance K atoms

Coefficient of \(KF\) is \(6\) (from Step 2), number of K atoms from \(KF\) is \(6\). Let the coefficient of \(K_3PO_3\) be \(w\). For K balance: \(3w=6\), so \(w = 2\).

Step4: Check PO₃ group balance

Coefficient of \(K_3PO_3\) is \(2\), so number of \(PO_3\) groups is \(2\). Coefficient of \(Ba_3(PO_3)_2\) is \(1\), number of \(PO_3\) groups is \(2\). So it is balanced.

Answer:

\(3BaF_2+2K_3PO_3
ightarrow6KF + Ba_3(PO_3)_2\)