Question

250 grams of silver nitrate react with an excess amount of potassium chloride to produce potassium nitrate and silver chloride. How many grams of silver chloride do you get?

Answer 1

Let's start by writing the balanced reaction equation:

AgNO₃ + KCl $\rightarrow$ AgCl$\downarrow$ + KNO₃.

As you can see, when one mole of AgNO₃ reacts, 1 mole of silver chloride is produced:

$n(AgNO_3) = n(AgCl)$ .

The number of the moles is the ratio of the mass $m$ to the molar mass $M$. The molar mass of AgNO₃ is 107.87+14.01+16.00·3=169.87 g/mol. The molar mass of AgCl is: 107.87+35.45=143.32 g/mol. The mass of silver chloride can be expressed as:

$m(AgCl) = n(AgCl)·M(AgCl)$

$m(AgCl) = n(AgNO_3)·M(AgCl)$

$m(AgCl) = \frac{m(AgNO_3)}{M(AgNO_3)}·M(AgCl)$

$m(AgCl) = \frac{250}{169.87}·143.32 = 210.9$ g.

Answer: When 250 grams of silver nitrate react with an excess amount of potassium chloride to produce potassium nitrate and silver chloride, you get 210.9 g of silver chloride.