# Question b7989

##### 2 Answers
Mar 25, 2015

0.02 moles are present.

1 mole of $C a C {l}_{2}$ weighs [40 + (35.5)2] = 111g

So 2g contains $\frac{2}{111} = 0.018 \text{mol}$

So we can see from the formula that the number of moles of $C {a}^{2 +}$ must also be equal to 0.018$\rightarrow 0.02 \text{mol}$

Mar 25, 2015

Your original solution contains $\text{0.02}$ moles of $C {a}^{2 +}$ cations.

So, you've got your double replacement reaction that culminates with the formation of a precipitate, calcium carbonate, $C a C {O}_{3}$.

The balanced chemical equation looks like this

$C a C {l}_{2 \left(a q\right)} + {K}_{2} C {O}_{3 \left(a q\right)} \to 2 K C {l}_{\left(a q\right)} + C a C {O}_{3 \left(s\right)}$

The complete ionic equation can be used to see what takes place between the ions present in solution

$C {a}_{\left(a q\right)}^{2 +} + 2 C {l}_{\left(a q\right)}^{-} + 2 {K}_{\left(a q\right)}^{+} + C {O}_{3 \left(a q\right)}^{2 -} \to 2 {K}_{\left(a q\right)}^{+} + 2 C {l}_{\left(a q\right)}^{-} + C a C {O}_{3 \left(s\right)}$

If you eliminate spectator ions, which are ions present on both sides of the reaction, you'll get the net ionic equation

$C {a}_{\left(a q\right)}^{2 +} + C {O}_{3 \left(a q\right)}^{2 -} \to C a C {O}_{3 \left(s\right)}$

The important thing to notice here is that calcium chloride dissociates in aqueous solution to form

$C a C {l}_{2 \left(a q\right)} r i g h t \le f t h a r p \infty n s C {a}_{\left(a q\right)}^{2 +} + 2 C {l}_{\left(a q\right)}^{-}$

Notice that 1 mole of calcium chloride dissociates into 1 mole of $C {a}^{2 +}$ cations and 2 moles of chloride anions, $C {l}^{-}$.

This means that the number of moles of $C {a}^{2 +}$ cations will be equal to the number of moles of calcium chloride. Therefore,

2cancel("g") CaCl_2 * "1 mole CaCl"_2/(110.98cancel("g")) = "0.018 moles CaCl"_2#

${\text{0.018"cancel("moles CaCl"_2) * ("1 mole Ca"^(2+))/(1cancel("mole CaCl"_2)) = "0.018 moles Ca}}^{2 +}$

Rounded to one sig fig, the number of sig figs given for 2 g, the answer will be

${n}_{C {a}^{2 +}} = \textcolor{red}{\text{0.02 moles}}$