Question

g For the following reaction, 0.500 moles of silver nitrate are mixed with 0.285 moles of copper(II) chloride. What is the formula for the limiting reagent? What is the maximum amount of silver chloride that can be produced?

Answer 1

Answer:

$CuCl_2$ is the formula for the limiting reagent.

Mass of silver chloride produced is 71.8 g.

Explanation:

$CuCl_2+2AgNO_3\rightarrow 2AgCl+Cu(NO_3)_2$

Moles of silver nitrate = 0.500 mol

Moles of copper(II) chloride = 0.285 mol

According to reaction, 2 moles of silver nitrate reacts with 1 mole of copper chloride , then 0.500 mole of silver nitrate will react with :

$\frac{1}{2}\times 0.500 mol=0.250 mol$ of copper(II) chloride

As we can see that moles of copper(II) chloride will be reacting is 0.250 mol less than present moles of copper (II) chloride ,so this means that silver nitrate is limiting reagent.

And moles of silver chloride to be formed will depend upon silver nitrate.

According to reaction, 2 moles of silver nitrate gives 2 moles of silver chloride , then 0.500 mole of silver nitrate will give  :

$\frac{2}{2}\times 0.500 mol=0.500 mol$ of silver chloride

Mass of silver chloride produced:

0.500 mol × 143.5 g/mol = 71.8 g