Jeremy R. answered • 09/23/19

Friendly, Experienced and Ivy League-trained science and math tutor

Let's start out by writing what we know.

a. Our chemical reaction: CuCl_{2(aq)}+ 2 KOH_{(aq)} --> Cu(OH)_{2(aq)} + 2 KCl_{(aq)}

b. What we have: 78.0 mL of 0.735 M CuCl_{2}

- We want to find out how much of a product chemical is formed from a certain amount of reactant chemical. And the only way we do that is by finding how many moles there are of the reactant. So to do that, let's recall that the molarity, M, of an aqueous solution is how many moles of a chemical there are in the solution. That is,
**M = mol/L**.

0.735 M CuCl_{2} = 0.735 mol CuCl_{2}/L

Now we can use dimensional analysis to find how many moles of CuCl_{2 }there are :

- 78.0
~~mL CuCl~~_{2}(1 L CuCl_{2}/ 1000~~mL CuCl~~_{2}) = 0.0780 L CuCl_{2} - 0.0780
~~L CuCl~~_{2}(0.735 mol CuCl_{2}/ 1~~L CuCl~~_{2}) =**0.05733 mol CuCl**_{2}

- Now let's find how many moles of Cu(OH)
_{2}that 0.05733 mol CuCl_{2}makes. Looking at our reaction above, we see that for every 1 mole of CuCl_{2}, we make 1 mole of Cu(OH)_{2}(Hint: look at the numbers in front of each chemical in the reaction -- that tells you the number of moles). So now we use dimensional analysis: - 0.05733
~~mol CuCl~~_{2}(1 mol Cu(OH)_{2}/ 1~~mol CuCl~~_{2}) =**0.05733 mol Cu(OH)**_{2}<-- It's the same because there's a 1:1 ratio between the reactant and the product.

- So now that we know how many moles of Cu(OH)
_{2}are made, we need to find how many grams that converts to. We first need to know the molar mass of Cu(OH)_{2}. And to do that, we add the atomic weights of every atom in the molecule, which you can find on a periodic table:

(1 x Cu) + (2 x O) + (2 x H) = (1 x 63.546 g) + (2 x 15.999 g) + (2 x 1.008 g)

= **97.546 g** **per mole of Cu(OH)**_{2}

Now we can solve our problem, again using dimensional analysis:

- 0.05733
~~mol Cu(OH)~~_{2}(_{ }97.546 g Cu(OH)_{2}/1~~mol Cu(OH)~~_{2}) =__5.59 g Cu(OH)___{2}__will precipitate out of solution__<-- This is the final answer, rounded to three significant figures because*each of our initial values (78.0 mL of 0.735 M CuCl*_{2}**)***had only three significant figures.*