Question

What mass (g) of magnesium nitride (Mg3N2) can be made from the reaction of 1.22 g of magnesium with excess nitrogen? __Mg + __N2__Mg3N2
a. 1.69 b. 15.2
c. 5.07 d. 5.02 e. 3.38
I’ll give brainest

Answer 1

Answer:

1.69 g Mg₃N₂

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

Atomic Structure

• Reading a Periodic Table

Stoichiometry

• Using Dimensional Analysis
• Reactions RxN

Explanation:

Step 1: Define

[RxN - Unbalanced] Mg + N₂ → Mg₃N₂

[RxN - Balanced] 3Mg + N₂ → Mg₃N₂

[Given] 1.22 g Mg

[Solve] grams Mg₃N₂

Step 2: Identify Conversions

[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

Step 3: Stoich

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]:                                                         $\displaystyle 1.68873 \ g \ Mg_3N_2$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂