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

1.69 g Mg₃N₂

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

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

Step 4: Check

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

1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year

hi my dude or dudette