2AgNO3 + BaCl2 → 2AgCl + Ba(NO3)2 How many grams of AgCl, silver chloride, are produced from 5.0 g of AgNO3, silver nitrate?

4 g AgCl

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Stoichiometry

Step 1: Define

[RxN]   2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

[Given]   5.0 g AgNO₃

Step 2: Identify Conversions

[Reaction - Stoich] 2AgNO₃ → 2AgCl

Molar Mass of Ag - 107.87 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

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

Step 3: Stoichiometry

Step 4: Check

Follow sig fig rules and round. We are given 1 sig fig.

4.21533 g AgCl ≈ 4 g AgCl

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