Asked by Emily
A sample of ammonium salt (NH4X) having a mass of 1.618g was heated with NaOH and the liberated ammonia was reacted with 50.00 of .6230 mol/L HCl. The excess acid required 1.50 mL of .6115 mol/L NaOH in titration. What is the formula of the salt?
Answers
Answered by
DrBob222
mols HCl used = 0.05000 x 0.6230 = ?
back titrated with NaOH = 0.0015 x 0.6115 = ?
mols NH3 present = total mols HCl-mols NaOH in back titration.
Then mol = grams/molar mass.
You know mols and grams, solve for molar mass. Subtract total molar mass - N - 4*H to find atomic mass of X then compare wth periodic table to identify X. I think it's Cl.
back titrated with NaOH = 0.0015 x 0.6115 = ?
mols NH3 present = total mols HCl-mols NaOH in back titration.
Then mol = grams/molar mass.
You know mols and grams, solve for molar mass. Subtract total molar mass - N - 4*H to find atomic mass of X then compare wth periodic table to identify X. I think it's Cl.
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