Question

Someone answer the limiting regent plz...

Answer 1

Answer:

$\boxed {\boxed {\sf About \ 16 \ grams \ of \ silver}}$

Explanation:

We are given the reaction:

$Cu_{s}+AgNO_{3(aq)} \rightarrow CuNO_{3(aq)}+Ag_{s}$

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.

• Silver (Ag): 107.868 g/mol
• Nitrogen (N) : 14.007 g/mol
• Oxygen (O): 15.999 g/mol

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

• O₃= (15.999 g/mol) *3=49.997 g/mol

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.

1. Convert grams of silver nitrate to moles.

$25 \ g \ AgNO_3*\frac{1 \ mol \ AgNO_3}{169.872 \ g \ AgNO_3}$

$\frac{25 \ mol \ AgNO_3}{169.872}= 0.1471696336 \ mol \ AgNO_3$

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

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

$0.1471696336 \ mol \ AgNO_3*\frac{1 \ mol \ Ag}{1 \ mol \ AgNO_3}=0.1471696336 \ mol \ Ag$

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

   3. Convert moles of silver to grams of silver.

$0.1471696336 \ mol \ Ag*\frac{107.868 \ g\ Ag}{1 \ mol \ Ag }$

$0.1471696336 *{107.868 \ g\ Ag}=15.87489404 \ g\ Ag$

   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.

$15.87489404 \ g \ Ag \approx 16 \ g\ Ag$

About 16 grams of silver would be produced.

Answer 2

Answer:

16 g Ag

General Formulas and Concepts:

Chemistry - Stoichiometry

• Using Dimensional Analysis

Chemistry - Atomic Structure

• Reading a Periodic Table

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

$25 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3} )(\frac{1 \ mol \ Ag}{1 \ mol \ AgNO_3} )(\frac{107.87 \ g \ Ag}{1 \ mol \ Ag} )$ = 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