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LuckyWell [14K]
3 years ago
13

Which of the following is a balanced chemical equation for the synthesis of NaBr from Na and Br2? A. 2Na + Br2 --> 2NaBr B. N

aBr --> Na + Br2 C. 2NaBr --> 2Na + Br 2 D. 2Na + Br2 → NaBr
Chemistry
2 answers:
max2010maxim [7]3 years ago
6 0
A. 2Na + Br2-->2NaBr.
goldfiish [28.3K]3 years ago
4 0

Answer:

A) 2Na + Br_{2} = 2NaBr

Explanation:

As the problem says that you should find the balanced chemical equation for the synthesis of NaBr from Na and Br_{2}, the reactants are Na and Br_{2} and the product is NaBr, so it can be writen as:

Na + Br_{2} = NaBr

Then you shoud balance the equation, it means that you should have the same number of each element in each part of the equation, so if you put a number 2 for the NaBr and the Na, the equation will be balanced:

2Na + Br_{2} = 2NaBr

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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
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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

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

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
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<h3>Explanation:</h3>

<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>

  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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