Question

2agno₃ + cacl₂ → 2agcl + ca(no₃)₂
you found that agno₃ is the limiting reactant. when 80.00 g agno₃ react, what mass of agcl forms?
? g agcl

Explanation:

Step1: Calculate moles of \( \text{AgNO}_3 \)

Molar mass of \( \text{AgNO}_3 \): \( \text{Ag} (107.87) + \text{N} (14.01) + 3\times\text{O} (16.00) = 107.87 + 14.01 + 48.00 = 169.88 \, \text{g/mol} \)
Moles of \( \text{AgNO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{80.00 \, \text{g}}{169.88 \, \text{g/mol}} \approx 0.471 \, \text{mol} \)

Step2: Relate moles of \( \text{AgNO}_3 \) to \( \text{AgCl} \)

From the balanced equation: \( 2 \, \text{mol} \, \text{AgNO}_3
ightarrow 2 \, \text{mol} \, \text{AgCl} \) (mole ratio \( 1:1 \))
Thus, moles of \( \text{AgCl} = \) moles of \( \text{AgNO}_3 = 0.471 \, \text{mol} \)

Step3: Calculate mass of \( \text{AgCl} \)

Molar mass of \( \text{AgCl} \): \( \text{Ag} (107.87) + \text{Cl} (35.45) = 143.32 \, \text{g/mol} \)
Mass of \( \text{AgCl} = \text{moles} \times \text{molar mass} = 0.471 \, \text{mol} \times 143.32 \, \text{g/mol} \approx 67.5 \, \text{g} \)

Answer:

\( \boxed{67.5} \) (Note: More precise calculation: \( \frac{80.00}{169.88} \times 143.32 \approx 67.48 \approx 67.5 \, \text{g} \))