Al2(s04)3 solution of a molal concentration is present in 1 litre solution of 2.684g/cc. How many moles of bas04 would be precipitated on adding bacl2 in excess?
the answer 6M
Who helps me step by step for me clearly
I am thankful to all tutors
To find the answer, we need to follow these steps:
Step 1: Convert the given density from grams per cubic centimeter (g/cc) to grams per liter (g/L).
Given density = 2.684 g/cc
Density in g/L = Density in g/cc * 1000 cc/L
Density in g/L = 2.684 g/cc * 1000 cc/L = 2684 g/L
Step 2: Calculate the molar mass of Al2(SO4)3 and BaSO4.
The molar mass of Al2(SO4)3 = (2 * Atomic mass of Al) + (3 * Atomic mass of S) + (12 * Atomic mass of O)
The molar mass of BaSO4 = Atomic mass of Ba + Atomic mass of S + (4 * Atomic mass of O)
You can look up the atomic masses of each element on the periodic table.
Step 3: Calculate the number of moles of Al2(SO4)3 in the 1 liter solution.
Number of moles = Mass of Al2(SO4)3 / Molar mass of Al2(SO4)3
Given mass of Al2(SO4)3 = Not provided
Without the given mass, we cannot calculate the number of moles of Al2(SO4)3, so we need more information to proceed.
Step 4: Calculate the number of moles of BaCl2 used in excess.
Since it is mentioned that BaCl2 is used in excess, it means that all the sulfate ions (SO4^2-) from Al2(SO4)3 will react with Ba^2+ ions to form BaSO4.
To calculate the number of moles of BaCl2 used in excess, we need to know the molar concentration of Al2(SO4)3 solution.
Step 5: Calculate the number of moles of BaSO4 precipitated.
Once we know the number of moles of BaCl2 used in excess, we can assume that all the sulfate ions react to form BaSO4. Therefore, the number of moles of BaSO4 precipitated will be equal to the number of moles of sulfate ions present in Al2(SO4)3.
Without the provided mass of Al2(SO4)3 or the molar concentration of Al2(SO4)3 solution, we cannot calculate the exact number of moles of BaSO4 precipitated on adding BaCl2 in excess. Therefore, we cannot determine if the answer is 6M or not.
Al2(SO4)3 +3BaCl2 >> 3BaSO4 (s) + 2AlCl3
first, figure out the number of moles per liter of aluminum sulfate there are in 2.684Kg.
You will get three times that number of moles of BaSO4 (look at the balanced equation).