Question

How many grams of silver chloride can be produced by reacting excess silver nitrate with 2.4 moles of zinc chloride? _____AgNO3 + ____ZnCl2  ____AgCl + _____Zn(NO3)2

Answer 1

690 g AgCl

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

Atomic Structure

• Reading a Periodic Table
• Writing Compounds

Stoichiometry

• Using Dimensional Analysis
• Limiting Reactant/Excess Reactant

Step-by-step explanation:

Step 1: Define

[RxN - Unbalanced] AgNO₃ + ZnCl₂ → AgCl + Zn(NO₃)₂

↓

[RxN - Balanced] 2AgNO₃ + ZnCl₂ → 2AgCl + Zn(NO₃)₂

[Given] 2.4 mol ZnCl₂

[Solve] x g AgCl

Step 2: Identify Conversions

[RxN] 1 mol ZnCl₂ → 2 mol AgCl

[PT] Molar Mass of Ag - 107.87 g/mol

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol

Step 3: Stoich

1. [DA] Set up:
   `\displaystyle 2.4 \ mol \ ZnCl_2((2 \ mol \ AgCl)/(1 \ mol \ ZnCl_2))((143.32 \ g \ AgCl)/(1 \ mol \ AgCl))`
2. [DA] Multiply/Divide [Cancel out units]:
   `\displaystyle 687.936 \ g \ AgCl`

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

687.936 g AgCl ≈ 690 g AgCl