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Chemistry 11 Online
OpenStudy (anonymous):

There are ___ mol of chloride ions in 0.500 L of a 0.300M solution of AlCl3 I know the first step is 0.300 x 0.500 = 0.150 mol and that the answer is 0.450

OpenStudy (ipwnbunnies):

There are 0.150 moles of the AlCl3 solution. But you need to use stoichiometry to mind the moles of chlorine in it.

OpenStudy (ipwnbunnies):

0.150 moles of AlCl3*

OpenStudy (anonymous):

So how exactly do I do that @iPwnBunnies ?

OpenStudy (ipwnbunnies):

You should know how to do that by now. As you can see in AlCl3, there are 3 moles of Cl for every 1 mole of AlCl3. So, 0.150 mol AlCl3 * (3 mol Cl per mol AlCl3)

OpenStudy (anonymous):

Okay. Got it.

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