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

690 g AgCl

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right  

Chemistry

Atomic Structure

Reading a Periodic TableWriting Compounds

Stoichiometry

Using Dimensional AnalysisLimiting Reactant/Excess ReactantExplanation:

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

[DA] Set up:                                                                                                      [DA] Multiply/Divide [Cancel out units]:                                                          

Step 4: Check

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

687.936 g AgCl ≈ 690 g AgCl

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