Question

Someone answer the limiting regent plz...

Answer 1

16 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] 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((1 \ mol \ AgNO_3)/(169.88 \ g \ AgNO_3) )((1 \ mol \ Ag)/(1 \ mol \ AgNO_3) )((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

Answer 2

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

Step-by-step 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*(1 \ mol \ AgNO_3)/(169.872 \ g \ AgNO_3)\\`

`(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*(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*(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.