Question

Question #b7989

Answer 1

0.02 moles are present.

1 mole of `CaCl_2` weighs [40 + (35.5)2] = 111g

So 2g contains `2/111= 0.018"mol"`

So we can see from the formula that the number of moles of `Ca^(2+)` must also be equal to 0.018`rarr0.02"mol"`

Answer 2

Your original solution contains `"0.02"` moles of `Ca^(2+)` cations.

So, you've got your double replacement reaction that culminates with the formation of a precipitate, calcium carbonate, `CaCO_3`.

The balanced chemical equation looks like this

`CaCl_(2(aq)) + K_2CO_(3(aq)) -> 2KCl_((aq)) + CaCO_(3(s))`

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

`Ca_((aq))^(2+) + 2Cl_((aq))^(-) + 2K_((aq))^(+) + CO_(3(aq))^(2-) -> 2K_((aq))^(+) + 2Cl_((aq))^(-) + CaCO_(3(s))`

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

`Ca_((aq))^(2+) + CO_(3(aq))^(2-) -> CaCO_(3(s))`

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

`CaCl_(2(aq)) rightleftharpoons Ca_((aq))^(2+) + 2Cl_((aq))^(-)`

Notice that 1 mole of calcium chloride dissociates into 1 mole of `Ca^(2+)` cations and 2 moles of chloride anions, `Cl^(-)`.

This means that the number of moles of `Ca^(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`

`"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_(Ca^(2+)) = color(red)("0.02 moles")`