Answer:
0.6 g NaCl
General Formulas and Concepts:
Math
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Chemistry
Atomic Structure
Stoichiometry
- Using Dimensional Analysis
Step-by-step explanation:
Step 1: Define
[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)
[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂
Step 2: Identify Conversions
[RxN] Na₂CO₃ → 2NaCl
Molar Mass of Na - 22.99 g/mol
Molar Mass of C - 12.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol
Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol
Step 3: Stoichiometry
- Set up:
- Multiply/Divide:
Step 4: Check
Follow sig fig rules and round. We are given 1 sig fig.
0.551373 g NaCl ≈ 0.6 g NaCl