Question

How many grams do you have in: 5.3 moles of Zr(NO3)4​

Answer 1

1800 g Zr(NO₃)₄

General Formulas and Concepts:

Chemistry - Atomic Structure

• Reading a Periodic Table
• Using Dimensional Analysis

Step-by-step 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

`5.3 \ mol \ Zr(NO_3)_4((339.26 \ g\ Zr(NO_3)_4)/(1 \ mol \ Zr(NO_3)_4) )` = 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