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Question

In a precipitation reaction between Al(NO₃)₃(aq) and Na₂S(aq), 12.6 mL of 0.150 M Al(NO₃)₃(aq) completly reacted with 10.8 mL of Na₂S(aq). What was the molarity of Na₂S(aq)?

2 Al(NO₃)₃(aq) + 3 Na₂S(aq)  → Al₂S₃(s)  + 6 NaNO₃(aq)

Note: Insert only the numerical value of your answer with three decimal places (do not include the units or chemical in your answer).

Answer 1

From the balanced equation we know that for every 2 moles of Al(NO₃)_3, you need 3 moles Na₂S to completely react the Al(NO₃)_3.

We can find the # of moles used in the reaction from the given information. We can use the concentration (molarity) of 0.150M and the amount used (12.6mL) to figure out how many moles of Al(NO₃)₃ was in the solution.

M = moles/1L of solution

So there are 0.150moles of Al(NO₃)₃ in 1 liter of the solution, however, we didn't use 1 L, we used 12.6mL, so the # of moles used was actually way less.

If 0.150 moles is equal to 1L, then how many moles (x) is equal to 12.6mL (0.0126L)? (Cross multiply)

0.150 = 1L

x = 0.0126L

x = 0.00189 moles of Al(NO₃)₃ was used in the reaction.

Remember, for every 2 moles of Al(NO₃)₃, we need 3 moles of Na₂S. (Understanding ratios can be tricky, so if you need an explanation please ask.) To find how many moles of Na₂S is needed given the ratio, we do:

(3/2) x 0.00189 = 0.002835 moles of Na₂S needed.

We are so close to the answer! But the question is asking for the molarity of the Na₂S used not the # of moles. So our answer needs to be in moles/L (M). We have the number of moles (0.002835), and the question says we used 10.8mL (0.0108L), thus we do:

0.002835 moles / 0.0108L = 0.2625 M

Answer: 0.2625 M Na₂S

The question wants only 3 decimal places: 0.263