Question #b7989

2 Answers
Mar 25, 2015

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 0.018rarr0.02"mol"

Mar 25, 2015

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")