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Answer:
a) The chemical reaction is given as:
2.571 grams of aluminum hydroxide is precipitated.
Explanation:
a) The chemical reaction is given as:
b)
Moles of NaOH = n
Volume of the NaOH = 185.5 mL = 0.1855 L( 1 mL =0.001 L)
Molarity of the solution = 0.533 M
Moles of aluminum = n
15.8 g of aluminum sulfate per liter.
Volume of solution = 627 mL = 0.627 L (1 mL= 0.001 L)
Mass of aluminium sulfate in solution = 15.8 g/L × 0.627 L =9.9066 g
Moles of aluminum sulfate =
Moles of NaOH = 0.09887 mol
According to reaction, 6 moles of NaOH reacts with 1 mole of aluminum sulfate, then 0.09887 moles of NaOH will recat with :
This means that sodium hydroxide moles are in limiting amount.So, amount of aluminum hydroxide will depend upon moles of sodium hydroxide.
According to reaction, 6 moles of sodium hydroxide gives 2 moles of aluminium hydroxide, then 0.09887 moles of sodium hydroxide will give :
of aluminum hydroxide
Mass of 0.03296 moles of aluminum hydroxide:
0.03296 mol × 78 g/mol = 2.571 g
2.571 grams of aluminum hydroxide is precipitated.