How many grams of ammonium fluoride are required to neutralize 972 'm of 0.00327 M sulfuric acid
To determine the grams of ammonium fluoride required to neutralize the sulfuric acid, we will first need to calculate the number of moles of sulfuric acid present. Then, using the balanced equation between sulfuric acid and ammonium fluoride, we can find the molar ratio between the two compounds. Finally, we will multiply the number of moles of sulfuric acid by the molar mass of ammonium fluoride to find the grams required.
Step 1: Calculate moles of sulfuric acid.
Moles = concentration (M) x volume (L)
Moles = 0.00327 mol/L x 0.972 L
Moles = 0.00317 mol
Step 2: Determine the molar ratio between sulfuric acid and ammonium fluoride.
From the balanced equation: H2SO4 + 2NH4F -> (NH4)2SO4 + 2HF
The molar ratio between sulfuric acid and ammonium fluoride is 1:2. This means that for every 1 mole of sulfuric acid, we need 2 moles of ammonium fluoride.
Step 3: Calculate grams of ammonium fluoride.
Moles of ammonium fluoride = 2 x moles of sulfuric acid
Moles of ammonium fluoride = 2 x 0.00317 mol
Moles of ammonium fluoride = 0.00634 mol
Grams of ammonium fluoride = moles of ammonium fluoride x molar mass of ammonium fluoride
Grams of ammonium fluoride = 0.00634 mol x (NH4F molecular weight)
The molar mass of ammonium fluoride (NH4F) is 53.49 g/mol.
Therefore, the grams of ammonium fluoride required to neutralize the sulfuric acid is approximately 0.338 grams (rounded to three decimal places).