Answer:
1800 g Zr(NO₃)₄
General Formulas and Concepts:
Chemistry - Atomic Structure
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
Step 1: Define
5.3 mol Zr(NO₃)₄
Step 2: Identify Conversions
Molar Mass of Zr - 91.22 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Zr(NO₃)₄ - 91.22 + 4(14.01) + 12(16.00) = 338.26 g/mol
Step 3: Convert
[tex]5.3 \ mol \ Zr(NO_3)_4(\frac{339.26 \ g\ Zr(NO_3)_4}{1 \ mol \ Zr(NO_3)_4} )[/tex] = 1798.08 g Zr(NO₃)₄
Step 4: Check
We are given 2 sig figs. Follow sig fig rules and round.
1798.08 g Zr(NO₃)₄ ≈ 1800 g Zr(NO₃)₄
