Answer:
1) 41.16 % = 0.182 grams
2) The alkali cation is K+ , to form the salt K2SO4
Explanation:
Step 1: Data given
Mass of unknown sulfate salt = 0.323 grams
Volume of water = 50 mL
Molarity of HCl = 6M
Step 2: The balanced equation
SO4^2- + BaCl2 → BaSO4 + 2Cl-
Step 3: Calculate amount of SO4^2- in BaSO4
The precipitate will be BaSO4
The amount of SO4^2- in BaSO4 = (Molar mass of SO4^2-/Molar mass BaSO4)*100
%
The amount of SO4^2- in BaSO4 = (96.06 /233.38) * 100
= 41.16%
So in 0.443g of BaSO4 there will be 0.443 * 41.16 % = <u>0.182 grams</u>
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2. If it is assumed that the salt is an alkali sulfate determine the identity of the alkali cation.
The unknown sulphate salt has 0.182g of sulphate. This means the alkali cation has a weight of 0.323-0.182 = 0.141g grams
An alkali cation has a chargoe of +1; sulphate has a charge of -2
The formula will be X2SO4 (with X = the unknown alkali metal).
Calculate moles of sulphate
Moles sulphate = 0.182 grams (32.1 + 4*16)
Moles sulphate = 0.00189 moles
The moles of sulphate = 0.182/(32.1+16*4)
The moles of sulphate = 0.00189 moles
X2SO4 → 2X+ + SO4^2-
For 2 moles cation we have 1 mol anion
For 0.00189 moles anion, we have 2*0.00189 = 0.00378 moles cation
Calculate molar mass
Molar mass = mass / moles
Molar mass = 0.141 grams / 0.00378 grams
Molar mass = 37.3 g/mol
The closest alkali metal is potassium. (K2SO4 )
