Question

1. agno₂ + mgcl₂ → mg(no₂)₂ + agcl

Explanation:

Step1: Balance Ag atoms

On the left - hand side, we have \(AgNO_3\), and on the right - hand side, we have \(AgCl\). Let the coefficient of \(AgNO_3\) be \(x\) and the coefficient of \(AgCl\) be \(y\). For Ag atoms: \(x=y\).

Step2: Balance \(NO_3^{-}\) groups

Since the coefficient of \(Mg(NO_3)_2\) is related to the \(NO_3^{-}\) groups. If the coefficient of \(AgNO_3\) is \(2\), then the number of \(NO_3^{-}\) groups is \(2\). For the compound \(Mg(NO_3)_2\), if the coefficient of \(Mg(NO_3)_2\) is \(1\), the number of \(NO_3^{-}\) groups from \(Mg(NO_3)_2\) is \(2\) (because each \(Mg(NO_3)_2\) has \(2\) \(NO_3^{-}\) groups).

Step3: Balance Cl atoms

We have \(MgCl_2\) on the left - hand side. Since the coefficient of \(AgCl\) is \(2\) (from the balance of Ag atoms when \(AgNO_3\) coefficient is \(2\)), and each \(MgCl_2\) has \(2\) Cl atoms. So the coefficient of \(MgCl_2\) is \(1\) (because \(1\times2\) (from \(MgCl_2\)) \(=2\) (from \(2AgCl\))).

Answer:

\(2AgNO_3 + 1MgCl_2
ightarrow1Mg(NO_3)_2+2AgCl\)