Question

Someone answer the limited regent...

Answer 1

20 g Ag

General Formulas and Concepts:

Chemistry - Stoichiometry

• Using Dimensional Analysis

Chemistry - Atomic Structure

• Reading a Periodic Table

Step-by-step explanation:

Step 1: Define

[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)

[Given] 10 g Cu

Step 2: Identify Conversions

[RxN] 1 mol Cu = 1 mol Ag

Molar Mass of Cu - 63.55 g/mol

Molar Mass of Ag - 197.87 g/mol

Step 3: Stoichiometry

`10 \ g \ Cu((1 \ mol \ Cu)/(63.55 \ g \ Cu))((1 \ mol \ Ag)/(1 \ mol \ Cu) )((197.87 \ g \ Ag)/(1 \ mol \ Ag) )` = 16.974 g Ag

Step 4: Check

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

16.974 g Ag ≈ 20 g Ag