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\( \begin{array}{l} \text { Balance this equation. }\\ -_{-} \mathrm{SnO}_{2}+{ }_{-} \mathrm{H}_{2}-{ }_{--} \mathrm{Sn}+{ }_{-} \mathrm{H}_{2} \mathrm{O} \end{array} \)

Ask by Gonzalez Vargas. in the United States
Jan 23,2025

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Answer

The balanced equation is: \[ \mathrm{SnO}_2 + 2\,\mathrm{H}_2 \rightarrow \mathrm{Sn} + 2\,\mathrm{H}_2\mathrm{O} \]

Solution

Sure, let's balance the chemical equation step by step. **Unbalanced Equation:** \[ \mathrm{SnO}_2 + \mathrm{H}_2 \rightarrow \mathrm{Sn} + \mathrm{H}_2\mathrm{O} \] **Step 1: Balance the Tin (Sn) Atoms** - **Left Side:** 1 Sn atom in \(\mathrm{SnO}_2\) - **Right Side:** 1 Sn atom in \(\mathrm{Sn}\) Tin is already balanced. **Step 2: Balance the Oxygen (O) Atoms** - **Left Side:** 2 O atoms in \(\mathrm{SnO}_2\) - **Right Side:** 1 O atom in \(\mathrm{H}_2\mathrm{O}\) To balance oxygen, place a coefficient of 2 before \(\mathrm{H}_2\mathrm{O}\): \[ \mathrm{SnO}_2 + \mathrm{H}_2 \rightarrow \mathrm{Sn} + 2\,\mathrm{H}_2\mathrm{O} \] **Step 3: Balance the Hydrogen (H) Atoms** - **Left Side:** 2 H atoms in \(\mathrm{H}_2\) - **Right Side:** \(2 \times 2 = 4\) H atoms in \(2\,\mathrm{H}_2\mathrm{O}\) To balance hydrogen, place a coefficient of 2 before \(\mathrm{H}_2\): \[ \mathrm{SnO}_2 + 2\,\mathrm{H}_2 \rightarrow \mathrm{Sn} + 2\,\mathrm{H}_2\mathrm{O} \] **Balanced Equation:** \[ \boxed{\mathrm{SnO}_2 + 2\,\mathrm{H}_2 \rightarrow \mathrm{Sn} + 2\,\mathrm{H}_2\mathrm{O}} \] This equation now has the same number of each type of atom on both sides, ensuring it's balanced.

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To balance the equation, we start with the unbalanced form: \[ \text{SnO}_2 + \text{H}_2 \rightarrow \text{Sn} + \text{H}_2\text{O} \] First, we see that there are 2 oxygen atoms on the reactant side (from SnO2) and only 1 on the product side (in H2O). Hence, we need to add a coefficient of 2 in front of H2O: \[ \text{SnO}_2 + \text{H}_2 \rightarrow \text{Sn} + 2 \text{H}_2\text{O} \] Now, we have 2 hydrogen atoms in H2O, thus we will add a coefficient of 2 to H2: \[ \text{SnO}_2 + 2 \text{H}_2 \rightarrow \text{Sn} + 2 \text{H}_2\text{O} \] Now let's count again: we have 1 Sn, 2 O, and 4 H on the reactant side, and we have 1 Sn, 2 O, and 4 H on the product side. The equation is balanced! So, the balanced equation looks like this: \[ \text{SnO}_2 + 2 \text{H}_2 \rightarrow \text{Sn} + 2 \text{H}_2\text{O} \]

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