Someone answer the limiting regent plz...

16 g Ag

General Formulas and Concepts:

Chemistry - Stoichiometry

Chemistry - Atomic Structure

Explanation:

Step 1: Define

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

[Given]   25 g AgNO₃

Step 2: Identify Conversions

[RxN]   1 mol AgNO₃ = 1 mol Ag

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

Step 3: Stoichiometry

= 15.8744 g Ag

Step 4: Check

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

15.8744 g Ag ≈ 16 g Ag

Explanation:

We are given the reaction:

We know that there are 25 grams of silver nitrate, or AgNO₃. First, we must find the molar mass of silver nitrate.

Silver Nitrate (AgNO₃)

Identify the molar masses of each element in silver nitrate using the Peirodic Table.

Next, calculate the molar mass. There is a subscript of 3 after the O, so we must multiply oxygen's molar mass by 3.

AgNO₃= (107.868 g/mol) + (14.007 g/mol)+(49.997 g/mol)=169.872 g/mol

Silver

Next, use stoichometry to find the mass of the silver.

In this reaction, 1 mole of silver nitrate yields 1 mole of silver.

   2. Convert moles of silver nitrate to moles of silver.

We know that silver's molar mass is 107.868 grams per mole.

   3. Convert moles of silver to grams of silver.

   4. Round

The original measurement given had 2 siginficant figures (25= 2 and 5). Therefore, we must round to 2 sig figs or the nearest whole number for this problem.

The 8 in the tenth place tells us to round up to the nearest whole number.

About 16 grams of silver would be produced.