Question

agno₃ + be → ag + be(no₃)₂
which reaction shows the correctly balanced equation?
4agno₃ + 2be → 2be(no₃)₂ + 4ag
2agno₃ + 2be → be(no₃)₂ + 2ag
ag(no₃) + be → be(no₃) + ag
2agno₃ + be → be(no₃)₂ + 2ag

Explanation:

Step1: Count atoms on both sides

For the original unbalanced equation \(AgNO_3 + Be
ightarrow Ag + Be(NO_3)_2\), on the left - hand side (LHS): 1 \(Ag\), 1 \(NO_3\), 1 \(Be\). On the right - hand side (RHS): 1 \(Ag\), 2 \(NO_3\), 1 \(Be\).

Step2: Balance \(NO_3\) group

Since there are 2 \(NO_3\) groups on the RHS and 1 on the LHS, put a coefficient of 2 in front of \(AgNO_3\). The equation becomes \(2AgNO_3+Be
ightarrow Ag + Be(NO_3)_2\). Now for \(Ag\): LHS has 2 \(Ag\) (from \(2AgNO_3\)), RHS has 1 \(Ag\).

Step3: Balance \(Ag\) atoms

Put a coefficient of 2 in front of \(Ag\) on the RHS. The balanced equation is \(2AgNO_3 + Be
ightarrow Be(NO_3)_2+2Ag\).

Let's check each option:

• For \(4AgNO_3 + 2Be

ightarrow 2Be(NO_3)_2+4Ag\):

• \(Ag\): LHS \(4\), RHS \(4\)
• \(NO_3\): LHS \(4\), RHS \(4\)
• \(Be\): LHS \(2\), RHS \(2\). But it is more complicated than necessary.
• For \(2AgNO_3 + 2Be

ightarrow Be(NO_3)_2+2Ag\):

• \(Be\): LHS \(2\), RHS \(1\) (unbalanced).
• For \(Ag_2(NO_3)_2 + Be

ightarrow Be(NO_3)_2+Ag_2\): Incorrect formula for \(AgNO_3\) (should be \(AgNO_3\) not \(Ag_2(NO_3)_2\)).

• For \(2AgNO_3 + Be

ightarrow Be(NO_3)_2+2Ag\):

• \(Ag\): LHS \(2\), RHS \(2\)
• \(NO_3\): LHS \(2\), RHS \(2\)
• \(Be\): LHS \(1\), RHS \(1\)

Answer:

\(2AgNO_3 + Be
ightarrow Be(NO_3)_2+2Ag\) (the fourth option in the given list)