Question

how many grams of silver nitrate are necessary to react completely with 7.000 moles of copper? cu + 2 agno₃ → cu(no₃)₂ + 2 ag 37.43 g 1189 g 889.6 g 2368 g 2378 g

Explanation:

Step1: Determine mole ratio from reaction

The balanced reaction is \( \text{Cu} + 2\text{AgNO}_3
ightarrow \text{Cu(NO}_3\text{)}_2 + 2\text{Ag} \). The mole ratio of \( \text{Cu} \) to \( \text{AgNO}_3 \) is \( 1:2 \).

Step2: Calculate moles of \( \text{AgNO}_3 \)

Given moles of \( \text{Cu} = 7.000 \, \text{mol} \). Using the mole ratio, moles of \( \text{AgNO}_3 = 2 \times 7.000 = 14.000 \, \text{mol} \).

Step3: Find molar mass of \( \text{AgNO}_3 \)

Molar mass of \( \text{Ag} = 107.87 \, \text{g/mol} \), \( \text{N} = 14.01 \, \text{g/mol} \), \( \text{O} = 16.00 \, \text{g/mol} \).
Molar mass of \( \text{AgNO}_3 = 107.87 + 14.01 + 3\times16.00 = 169.88 \, \text{g/mol} \).

Step4: Calculate mass of \( \text{AgNO}_3 \)

Mass = moles × molar mass = \( 14.000 \, \text{mol} \times 169.88 \, \text{g/mol} = 2378.32 \, \text{g} \) (close to 2378 g).

Answer:

2378 g (Option: 2378 g)