Question

When 25.0 g of silver nitrate, AgNO₃, is reacted with 5.00 g Cu, how many grams of Ag can you theoretically produce?

Cu(s) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 Ag(s)

Question 6 options:

	32.9 g Ag
	15.9 g Ag
	17.0 g Ag
	5.00 g Ag

Answer 1

Answer 2

Balanced Chemical Equation:

$$\text{Cu (s)}+2{\text{AgNO}}_3\text{(aq)}\to {\text{Cu(NO}}_3{\text{)}}_2\text{(aq)}+2\text{Ag (s)}$$

Given:

• Mass of ${\text{AgNO}}_3=25.0\text{g}$

• Mass of $\text{Cu}=5.00\text{g}$

• Molar masses:

• ${\text{AgNO}}_3=169.87\text{g/mol}$

• $\text{Cu}=63.55\text{g/mol}$

• $\text{Ag}=107.87\text{g/mol}$

Objective:

Calculate the theoretical yield of $\text{Ag}$ in grams.

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Step 1: Convert Masses to Moles

1. Moles of ${\text{AgNO}}_3$:

   $$n_{{\text{AgNO}}_3}=\frac{25.0\text{g}}{169.87\text{g/mol}}=0.1472\text{mol}$$

2. Moles of $\text{Cu}$:

   $$n_{\text{Cu}}=\frac{5.00\text{g}}{63.55\text{g/mol}}=0.0787\text{mol}$$

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Step 2: Determine the Limiting Reactant

The balanced equation shows:

• 1 mol $\text{Cu}$ reacts with 2 mol ${\text{AgNO}}_3$.

1. Required ${\text{AgNO}}_3$ for all $\text{Cu}$:

   $$\text{Required }{\text{AgNO}}_3=2\times 0.0787\text{mol}=0.1574\text{mol}$$

• Available ${\text{AgNO}}_3=0.1472\text{mol}$ (less than required).

2. 
   

3. Conclusion:
   ${\text{AgNO}}_3$ is the limiting reactant because there isn’t enough to fully react with all the $\text{Cu}$.

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Step 3: Calculate Moles of $\text{Ag}$ Produced

From the equation:

• 2 mol ${\text{AgNO}}_3$ produces 2 mol $\text{Ag}$.

Thus, $0.1472\text{mol}$ of ${\text{AgNO}}_3$ produces:

$$n_{\text{Ag}}=0.1472\text{mol}$$

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Step 4: Convert Moles of $\text{Ag}$ to Grams

$$\text{Mass of Ag}=0.1472\text{mol}\times 107.87\text{g/mol}=15.87\text{g}$$

Answer:

15.9 g Ag (rounded to 3 significant figures).