1. _NaOH + _Cu(NO3)2 → _Cu(0H)2 (s) + _NaNO3| nen 25.00g of NaOH reacts with 25.00g Cu(NO3)2, how many moles of copper (II) hydroxide will produced?

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Balance the chemical reaction, and complete the stoichiometry problem attached. 

 

**Stoichiometry Problem**

**Balance the following chemical equation:**

\[ \_\_  \text{NaOH} +  \_\_ \text{Cu(NO}_3\text{)}_2 \rightarrow \_\_ \text{Cu(OH)}_2 \text{ (s) } + \_\_ \text{NaNO}_3 \]

**Problem:**

When 25.00 g of NaOH reacts with 25.00 g of Cu(NO\(_3\))\(_2\), how many moles of copper (II) hydroxide will be produced?

***Solution:***

**Step 1: Balance the chemical equation.**

First, write the unbalanced chemical equation:

\[ \text{NaOH} + \text{Cu(NO}_3\text{)}_2 \rightarrow \text{Cu(OH)}_2 + \text{NaNO}_3 \]

Now balance the equation by adjusting the coefficients. The balanced equation is:

\[ 2 \text{NaOH} + \text{Cu(NO}_3\text{)}_2 \rightarrow \text{Cu(OH)}_2 + 2 \text{NaNO}_3 \]

**Step 2: Determine the molar masses of the reactants.**

- Molar mass of NaOH: \(23.0 + 16.0 + 1.0 = 40.0 \, \text{g/mol}\)
- Molar mass of Cu(NO\(_3\))\(_2\): \(63.5 + 2 \times (14.0 + 3 \times 16.0) = 187.5 \, \text{g/mol}\)

**Step 3: Convert masses to moles.**

\[ \text{Moles of NaOH} = \frac{25.00 \, \text{g}}{40.0 \, \text{g/mol}} \approx 0.625 \, \text{moles} \]
\[ \text{Moles of Cu(NO}_3\text{)}_2 = \frac{25.00 \, \text{g}}{187.5 \, \text{g/mol}} \approx 0.133 \, \text{moles} \]

**Step 4: Determine the limiting reactant.**

From the balanced equation:
\[ 2
Transcribed Image Text:**Stoichiometry Problem** **Balance the following chemical equation:** \[ \_\_ \text{NaOH} + \_\_ \text{Cu(NO}_3\text{)}_2 \rightarrow \_\_ \text{Cu(OH)}_2 \text{ (s) } + \_\_ \text{NaNO}_3 \] **Problem:** When 25.00 g of NaOH reacts with 25.00 g of Cu(NO\(_3\))\(_2\), how many moles of copper (II) hydroxide will be produced? ***Solution:*** **Step 1: Balance the chemical equation.** First, write the unbalanced chemical equation: \[ \text{NaOH} + \text{Cu(NO}_3\text{)}_2 \rightarrow \text{Cu(OH)}_2 + \text{NaNO}_3 \] Now balance the equation by adjusting the coefficients. The balanced equation is: \[ 2 \text{NaOH} + \text{Cu(NO}_3\text{)}_2 \rightarrow \text{Cu(OH)}_2 + 2 \text{NaNO}_3 \] **Step 2: Determine the molar masses of the reactants.** - Molar mass of NaOH: \(23.0 + 16.0 + 1.0 = 40.0 \, \text{g/mol}\) - Molar mass of Cu(NO\(_3\))\(_2\): \(63.5 + 2 \times (14.0 + 3 \times 16.0) = 187.5 \, \text{g/mol}\) **Step 3: Convert masses to moles.** \[ \text{Moles of NaOH} = \frac{25.00 \, \text{g}}{40.0 \, \text{g/mol}} \approx 0.625 \, \text{moles} \] \[ \text{Moles of Cu(NO}_3\text{)}_2 = \frac{25.00 \, \text{g}}{187.5 \, \text{g/mol}} \approx 0.133 \, \text{moles} \] **Step 4: Determine the limiting reactant.** From the balanced equation: \[ 2
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