Question

do the following for questions 2-5: count the atoms of each element and write them under the reactants and products the picture. draw the atoms or molecules that are missing in the products, paying attention to the sizes of the atoms, then add the missing coefficient(s) to your chemical equation. 2. 2 nacl → na⁺ + 2 cl⁻ 3. sno₂ + h₂ → sn + 2 h₂o 4. n₂o₅ → 2 n₂ + o₂

Explanation:

Step1: Balance question 2 (Na atoms)

Reactants: $2$ NaCl has $2$ Na atoms. Products currently have $1$ Na⁺. Add coefficient 2 to Na⁺:
$2\text{NaCl}
ightarrow 2\text{Na}^+ + 2\text{Cl}^-$
Count check: Na: $2=2$, Cl: $2=2$

Step2: Verify question 3 (already balanced)

Reactants: $\text{SnO}_2 + 2\text{H}_2$ has 1 Sn, 2 O, 4 H.
Products: $\text{Sn} + 2\text{H}_2\text{O}$ has 1 Sn, $2\times1=2$ O, $2\times2=4$ H.
Atoms are balanced, coefficient 2 for $\text{H}_2$ is correct.

Step3: Balance question 4 (O atoms)

Reactants: $2\text{N}_2\text{O}_5$ has $2\times5=10$ O atoms.
Products: $2\text{N}_2 + x\text{O}_2$ has $2x$ O atoms. Solve $2x=10$ → $x=5$.
Balanced equation: $2\text{N}_2\text{O}_5
ightarrow 2\text{N}_2 + 5\text{O}_2$
Count check: N: $2\times2=4=2\times2$, O: $10=5\times2$

Answer:

1. Balanced equation: $\boldsymbol{2\text{NaCl}

ightarrow 2\text{Na}^+ + 2\text{Cl}^-}$ (add 1 more $\text{Na}^+$ particle to the products)

1. The given coefficient 2 for $\text{H}_2$ is correct, equation is balanced: $\boldsymbol{\text{SnO}_2 + 2\text{H}_2

ightarrow \text{Sn} + 2\text{H}_2\text{O}}$

1. Balanced equation: $\boldsymbol{2\text{N}_2\text{O}_5

ightarrow 2\text{N}_2 + 5\text{O}_2}$ (replace the 8 with 5, and draw 5 total $\text{O}_2$ molecules (10 O atoms total) in the products)