Determine the percent composition of each element in NH4Br. It should be rounded to the nearest percent. 2.) 3.25 x 10^25 atoms of neon gas (Ne) equals how many moles of neon gas (Ne)? 3.) How many moles of Al can be produced from 10.87g of Ag? Al(NO3)3(s) + 3Ag —> Al + 3AgNO3 4.) How many grams of N2 are needed to produce 2.17 mol of NH3 when reacted according to this chemical equation? N2(g) + 3H2(g) —> 2NH3(g)
The Correct Answer and Explanation is:
1. Percent Composition of NH₄Br
Molar masses:
- N = 14 g/mol
- H = 1 g/mol × 4 = 4 g/mol
- Br = 79.9 g/mol
Molar mass of NH₄Br = 14 + 4 + 79.9 = 97.9 g/mol
Percent Composition:
- N: (14 / 97.9) × 100 ≈ 14%
- H: (4 / 97.9) × 100 ≈ 4%
- Br: (79.9 / 97.9) × 100 ≈ 82%
Answer:
Nitrogen = 14%, Hydrogen = 4%, Bromine = 82%
2. Moles from Atoms of Neon
Use Avogadro’s number:
1 mole = 6.022 × 10²³ atoms
Given: 3.25 × 10²⁵ atoms Ne
Calculation:
Moles = (3.25 × 10²⁵ atoms) / (6.022 × 10²³ atoms/mol) ≈ 54.0 mol
Answer:
54 moles of Ne
3. Moles of Al from 10.87 g of Ag
Balanced equation:
Al(NO₃)₃ + 3Ag → Al + 3AgNO₃
Molar mass of Ag = 107.87 g/mol
Given mass = 10.87 g
Moles of Ag = 10.87 / 107.87 ≈ 0.101 mol
From stoichiometry:
3 mol Ag → 1 mol Al
So,
0.101 mol Ag × (1 mol Al / 3 mol Ag) ≈ 0.0337 mol Al
Answer:
0.034 moles of Al
4. Grams of N₂ to Produce 2.17 mol NH₃
Balanced equation:
N₂ + 3H₂ → 2NH₃
From stoichiometry:
2 mol NH₃ → 1 mol N₂
So,
2.17 mol NH₃ × (1 mol N₂ / 2 mol NH₃) = 1.085 mol N₂
Molar mass of N₂ = 28.02 g/mol
Mass = 1.085 mol × 28.02 g/mol ≈ 30.4 g
Answer:
30 grams of N₂
Explanation (Summary of Concepts Used)
These problems use key chemistry concepts: percent composition, mole conversions, and stoichiometry. Percent composition requires dividing each element’s mass by the total molar mass and multiplying by 100. Converting atoms to moles uses Avogadro’s number. Stoichiometry problems follow balanced chemical equations and molar ratios. For mass-to-mole or mole-to-mass conversions, the molar mass from the periodic table bridges between grams and moles. Each solution here ensures correct scientific methodology and proper rounding to the nearest whole number where needed.
