Question

2 NaOH (s) + CO2(g) → Na2CO3 (s) + H20 (I)

How many grams of water can be produced with 1.85 moles of NaOH​

Answer 1

Answer:

16.7 g H₂O

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

Stoichiometry

• Reading a Periodic Table
• Using Dimensional Analysis

Explanation:

Step 1: Define

[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)

[Given] 1.85 mol NaOH

Step 2: Identify Conversions

[RxN] 2 mol NaOH → 1 mol H₂O

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Stoichiometry

1. Set up:                               $\displaystyle 1.85 \ mol \ NaOH(\frac{1 \ mol \ H_2O}{2 \ mol \ NaOH})(\frac{18.02 \ g \ H_2O}{1 \ mol \ H_2O})$
2. Multiply/Divide:                 $\displaystyle 16.6685 \ g \ H_2O$

Step 4: Check

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

16.6685 g H₂O ≈ 16.7 g H₂O