30 cm³ of 0.1 M Al (NO₃)₃ SOLUTION IS RECTED WITH 100cm³ of 0.15M of NaOH solution. Which reactant is in excess, and by now much ?

Som15432

1 year ago

Let's figure this out step-by-step!
1. Find the moles of each reactant:
* Al(NO₃)₃:
* Moles = (Volume in liters) x (Molarity)
* Moles = (30 cm³ / 1000) x 0.1 M = 0.003 moles
* NaOH:
* Moles = (Volume in liters) x (Molarity)
* Moles = (100 cm³ / 1000) x 0.15 M = 0.015 moles
2. Use the balanced equation:
* The reaction is: Al(NO₃)₃ + 3NaOH → Al(OH)₃ + 3NaNO₃
* This means 1 mole of Al(NO₃)₃ reacts with 3 moles of NaOH.
3. Find out how much NaOH is needed:
* To react with 0.003 moles of Al(NO₃)₃, you need 0.003 moles x 3 = 0.009 moles of NaOH.
4. Determine the excess:
* You have 0.015 moles of NaOH, and you only need 0.009 moles.
* Excess NaOH = 0.015 moles - 0.009 moles = 0.006 moles.
5. Convert the excess moles of NaOH back to volume:
* Volume = Moles / Molarity
* Volume = 0.006 moles / 0.15M = 0.04 Liters
* 0.04 Liters * 1000 cm3/Liter = 40 cm3.
Therefore, the answer is C. NaOH solution by 40 cm³.

CenSmart

5 years ago

divive the volume by 1000