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djverab [1.8K]
2 years ago
7

How many grams of NaCl are contained in 0.50 L of a 4.20 M solution?

Chemistry
1 answer:
disa [49]2 years ago
8 0

Answer:

The answer to your question is the third option: 122.7 g of NaCl

Explanation:

Data

mass of NaCl = ?

molecular mass of NaCl = 23 + 35.5 = 58.5

volume = 0.50 l

concentration = 4.20 M

Formula

Molarity = \frac{moles}{volume}

     Moles = Molarity x volume

Process

1.- Substitute the data given to find the moles of NaCl

     Moles = 4.20 x 0.5

     Moles = 2.1

2.- Calculate the grams of NaCl using proportions

     58.5 g of NaCl ----------------- 1 mol

        x                     ----------------  2.1 moles

      x = (2.1 x 58.5) / 1

3.- Result

      x = 122.7 g of NaCl      

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What mass (g) of magnesium nitride (Mg3N2) can be made from the reaction of 1.22 g of magnesium with excess nitrogen? __Mg + __N
Citrus2011 [14]
<h3>Answer:</h3>

1.69 g Mg₃N₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
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<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

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<u>Step 1: Define</u>

[RxN - Unbalanced] Mg + N₂ → Mg₃N₂

[RxN - Balanced] 3Mg + N₂ → Mg₃N₂

[Given] 1.22 g Mg

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<u>Step 2: Identify Conversions</u>

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Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                      \displaystyle 1.22 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{1 \ mol \ Mg_3N_2}{3 \ mol \ Mg})(\frac{100.95 \ g \ Mg_3N_2}{1 \ mol\ Mg_3N_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 1.68873 \ g \ Mg_3N_2

<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

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