Answer: 67.50 g
Explanation:
Hi! In order to solve this question, we should use the moles of AgNO3 that react. This is because AgNO3 has the same molar coefficient as AgCl on the other side of the equation. This means that the moles of AgNO3 that are produced is the same as the number of moles of AgCl. From there we can multiply the mole value by AgCl's molar mass and get our answer.
Solving:
[tex]\ce{AgCl} (\text{silver chloride}): \\ \quad \ce{Ag} = 107.87 \, \text{g/mol}\\ \quad \ce{Cl} = 35.45 \, \text{g/mol} \\& \quad \text{Molar mass of} \, \ce{AgCl} = 107.87 + 35.45 = \boxed{143.32} \, \text{g/mol}[/tex]
[tex]\[\text{Moles of} \, \ce{AgNO3} = \frac{\text{Mass of} \, \ce{AgNO3}}{\text{Molar mass of} \, \ce{AgNO3}}\]\[\text{Moles of} \, \ce{AgNO3} = \frac{80.00 \, \text{g}}{169.88 \, \text{g/mol}} \approx \boxed{0.471} \, \text{moles}\][/tex]
Now we know that there are 0.471 moles of AgCl since their molar quantities are equal and all you have to do is multiply moles of AgCl by its molar mass:
[tex]\[\text{Mass of} \, \ce{AgCl} = \text{Moles of} \, \ce{AgCl} \times \text{Molar mass of} \, \ce{AgCl}\]\[\text{Mass of} \, \ce{AgCl} = 0.471 \, \text{moles} \times 143.32 \, \text{g/mol} \approx \boxed{\boxed{67.50}} \, \text{grams}\][/tex]
That's it!
