Asked by anon
Given a solution containing 0.242 g of aqueous barium chloride, BaCl2, how many mL of 0.0581 M H2SO4 solution will completely precipitate the barium ions, Ba2+? The molar mass of BaCl2 is 208.3 g/mol.
Answers
Answered by
anon
a) 20 mL
b) 18 mL
c) 24 mL
d) 30 mL
b) 18 mL
c) 24 mL
d) 30 mL
Answered by
DrBob222
BaCl2 + H2SO4 --> 2HCl + BaSO4
mols BaCl2 = grams/molar mass = 0.242/208 = 0.00116
From the equation, 1 mol BaCl2 = 1 mol H2SO4 so it will talke 0.00116 mols H2SO4. M H2SO4 = mols/L. You know M and mols, solve for L and convert to mL.
mols BaCl2 = grams/molar mass = 0.242/208 = 0.00116
From the equation, 1 mol BaCl2 = 1 mol H2SO4 so it will talke 0.00116 mols H2SO4. M H2SO4 = mols/L. You know M and mols, solve for L and convert to mL.
Answered by
M.
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