How many moles of HCl are needed to react with 0.87 mol Al?
1 answer:
You might be interested in
Answer:
The molar concentration would have to be 0,81 M.
Explanation:
The osmotic pressure equation is:
where:
: osmotic pressure [atm]
M: molar concentration [M]
R: gas constant 0,08205 [atm.L/mol.°K]
T: absolute temperature [°K]
To solve the problem, we just clear M from the osmotic pressure equation and then replace our data using the appropiate units. Clearing the variable M we have:
We have to use temperature as absolute temperature (in kelvins), T=29+273=302 °K. Now we can replace our values in the equation:
As we can see, all units will be simplified and we'll have the molar concentration in mol/L.
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.