Question

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Explanation:

Step1: Balance equation 7 (double displacement)

First, match cations/anions: $\text{Mg(NO}_3\text{)}_2 + 2\text{NaBr}
ightarrow \text{MgBr}_2 + 2\text{NaNO}_3$
Count atoms: Mg:1=1, N:2=2, O:6=6, Na:2=2, Br:2=2. Balanced.

Step2: Balance equation 8 (acid-base neutralization)

Match ions, balance $\text{NH}_4^+$ and $\text{OH}^-$: $\text{H}_2\text{SO}_4 + 2\text{NH}_4\text{OH}
ightarrow 2\text{H}_2\text{O} + (\text{NH}_4)_2\text{SO}_4$
Count atoms: H:2+8=10; 4+8=12? Correct: $\text{H}_2\text{SO}_4$ has 2 H, $2\text{NH}_4\text{OH}$ has 2*(4+1)=10 H → total 12 H. $2\text{H}_2\text{O}$ has 4 H, $(\text{NH}_4)_2\text{SO}_4$ has 8 H → 12 H. S:1=1, O:4+2=6; 2+4=6. N:2=2. Balanced.

Step3: Balance equation 9 (double displacement)

Rearrange reactants/products, balance Cl and P: $\text{H}_3\text{PO}_4 + 5\text{HCl}
ightarrow \text{PCl}_5 + 4\text{H}_2\text{O}$
Count atoms: H:3+5=8; 8=8. P:1=1. O:4=4. Cl:5=5. Balanced.

Answer:

1. $\boldsymbol{1\text{Mg(NO}_3\text{)}_2 + 2\text{NaBr}

ightarrow 1\text{MgBr}_2 + 2\text{NaNO}_3}$

1. $\boldsymbol{1\text{H}_2\text{SO}_4 + 2\text{NH}_4\text{OH}

ightarrow 2\text{H}_2\text{O} + 1(\text{NH}_4)_2\text{SO}_4}$

1. $\boldsymbol{1\text{H}_3\text{PO}_4 + 5\text{HCl}

ightarrow 1\text{PCl}_5 + 4\text{H}_2\text{O}}$

(Note: The coefficients 1 are optional but included for clarity; omitting them is also acceptable in balanced equations.)