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Answer:

23.459 g NaNO₂

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Chemistry

Stoichiometry

  • Reading a Periodic Table
  • Using Dimensional Analysis

Explanation:

Step 1: Define

[RxN] H₂SO₄ + 2NaNO₂ → 2HNO₂ + Na₂SO₄

[Given] 24.14714 g Na₂SO₄

Step 2: Identify Conversions

[RxN] 1 mol Na₂SO₄ = 2 mol NaNO₂

Molar Mass of Na - 22.99 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of Na₂SO₄ - 2(22.99) + 32.07 + 4(16.00) = 142.05 g/mol

Molar Mass of NaNO₂ - 22.99 + 14.01 + 2(16.00) = 69.00 g/mol

Step 3: Stoichiometry

  1. Set up:                              [tex]\displaystyle 24.14714 \ g \ Na_2SO_4(\frac{1 \ mol \ Na_2SO_4}{142.05 \ g \ Na_2SO_4})(\frac{2 \ mol \ NaNO_2}{1 \ mol \ Na_2SO_4})(\frac{69.00 \ g \ NaNO_2}{1 \ mol \ NaNO_2})[/tex]
  2. Multiply/Divide:                                                                                                [tex]\displaystyle 23.4587 \ g \ NaNO_2[/tex]

Step 4: Check

Follow sig fig rules and round. We need 5 sig figs (instructed).

23.4587 g NaNO₂ ≈ 23.459 g NaNO₂

Answer:

23.458

Explanation:

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