Answer:
152.4 g
Explanation:
Step 1: Given data
Number of H atoms: 8.334 × 10²⁴ atoms
Step 2: Calculate the moles of H corresponding to 8.334 × 10²⁴ atoms of H
We will use Avogadro's number, that is, there are 6.022 × 10²³ atoms of H in 1 mole of H.
8.334 × 10²⁴ atom × (1 mol/6.022 × 10²³ atom) = 13.84 mol
Step 3: Calculate the moles of ammonium acetate (NH₄CH₃CO₂) that contain 13.84 moles of H
The molar ratio of NH₄CH₃CO₂ to H is 1:7. The moles of NH₄CH₃CO₂ that contain 13.84 moles of H are 1/7 × 13.84 mol = 1.977 mol.
Step 4: Calculate the mass corresponding to 1.977 moles of NH₄CH₃CO₂
The molar mass of NH₄CH₃CO₂ is 77.08 g/mol.
1.977 mol × 77.08 g/mol = 152.4 g
