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<h3>Answer:</h3>
1.69 g Mg₃N₂
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right<u> </u>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Reactions RxN
<u>Step 1: Define</u>
[RxN - Unbalanced] Mg + N₂ → Mg₃N₂
[RxN - Balanced] 3Mg + N₂ → Mg₃N₂
[Given] 1.22 g Mg
[Solve] grams Mg₃N₂
<u>Step 2: Identify Conversions</u>
[RxN] 3 mol Mg → Mg₃N₂
[PT] Molar Mass of Mg - 24.31 g/mol
[PT] Molar Mass of N - 14.01 g/mol
Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol
<u>Step 3: Stoich</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂
Answer:
Concentration of the barium ions =
Concentration of the chloride ions =
Explanation:
Moles of hydrogen chloride = n
Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L
Molarity of the hydrogen chloride = 0.1355 M
According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.
Then 0.05947 moles of HCl will react with:
barium hydroxide
Moles of barium hydroxide = 0.029735 mol
1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:
Volume of solution after neutralization reaction :
= 20.0 mL + 43.89 mL = 63.89 mL = 0.06389 L
Concentration of the barium ions =
1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.
Then concentration of chloride ions will be: