mole test review: due
problems: for the following problems, show all work and report your answers with the correct units and significant figures.
1. what is the molar mass of baco₃?
2. in one molecule of p₂o₅, how many phosphorous atoms are present?
3. list all seven diatomics.
4. how many atoms of carbon are in 3.6 moles of carbon?
5. determine the number of moles of hydrogen that are in 362.8 g of this element.
6. a chemical reaction produces 13.8 mol of carbon monoxide gas. what volume will the gas occupy at stp?
7. what is the mass of 5.0 × 10²⁵ formula units of fe(no₂)₃?

Question

Accepted Answer

### Problem 4: How many atoms of carbon are in 3.6 moles of carbon?
# Explanation:
## Step 1: Recall Avogadro's Number
Avogadro's number is $ 6.022 \times 10^{23} $ atoms per mole, which relates moles to number of particles (atoms, in this case). The formula to find the number of atoms is: $ \text{Number of atoms} = \text{Moles} \times \text{Avogadro's Number} $
## Step 2: Substitute Values
Given moles of carbon ($ n $) = 3.6 mol, Avogadro's number ($ N_A $) = $ 6.022 \times 10^{23} $ atoms/mol.
So, $ \text{Number of C atoms} = 3.6 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} $
Calculate the product: $ 3.6 \times 6.022 \times 10^{23} \approx 2.17 \times 10^{24} $ atoms.
# Answer: $ 2.17 \times 10^{24} $ atoms of carbon

### Problem 5: Moles of hydrogen in 362.8 g of $ \text{H}_2 $
# Explanation:
## Step 1: Molar Mass of $ \text{H}_2 $
Molar mass of H is 1 g/mol, so for $ \text{H}_2 $, molar mass ($ M $) = $ 2 \times 1 = 2 \, \text{g/mol} $.
The formula for moles is $ n = \frac{m}{M} $, where $ m $ is mass, $ M $ is molar mass.
## Step 2: Substitute Values
Given $ m = 362.8 \, \text{g} $, $ M = 2 \, \text{g/mol} $.
$ n = \frac{362.8 \, \text{g}}{2 \, \text{g/mol}} = 181.4 \, \text{mol} $.
# Answer: $ 181.4 \, \text{moles} $ of $ \text{H}_2 $

### Problem 6: Volume of CO at STP
# Explanation:
## Step 1: STP Molar Volume
At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 L. The formula is $ V = n \times 22.4 \, \text{L/mol} $, where $ n $ is moles.
## Step 2: Substitute Values
Given $ n = 13.8 \, \text{mol} $ of CO.
$ V = 13.8 \, \text{mol} \times 22.4 \, \text{L/mol} = 309.12 \, \text{L} $.
# Answer: $ 309.12 \, \text{L} $ of CO

### Problem 7: Mass of $ 5.0 \times 10^{25} $ formula units of $ \text{Fe(NO}_2\text{)}_3 $
# Explanation:
## Step 1: Molar Mass of $ \text{Fe(NO}_2\text{)}_3 $
- Fe: 55.8 g/mol
- N: $ 3 \times 14 = 42 $ g/mol
- O: $ 6 \times 16 = 96 $ g/mol
Molar mass ($ M $) = $ 55.8 + 42 + 96 = 193.8 \, \text{g/mol} $ (approx 194 g/mol for calculation).

## Step 2: Moles from Formula Units
Use Avogadro's number: $ n = \frac{\text{Formula Units}}{N_A} $, where $ N_A = 6.02 \times 10^{23} \, \text{formula units/mol} $.
$ n = \frac{5.0 \times 10^{25}}{6.02 \times 10^{23}} \approx 83.06 \, \text{mol} $.

## Step 3: Mass from Moles
Mass ($ m $) = $ n \times M $.
$ m = 83.06 \, \text{mol} \times 194 \, \text{g/mol} \approx 1.61 \times 10^4 \, \text{g} $ (or using the given 194 g/mol: $ \frac{5.0 \times 10^{25}}{6.02 \times 10^{23}} \times 194 \approx 1.61 \times 10^4 \, \text{g} $).
# Answer: Approximately $ 1.61 \times 10^4 \, \text{g} $ (or $ 16100 \, \text{g} $) of $ \text{Fe(NO}_2\text{)}_3 $

### Problem 1: Molar Mass of $ \text{BaCO}_3 $
# Explanation:
## Step 1: Atomic Masses
- Ba: 137 g/mol
- C: 12 g/mol
- O: $ 3 \times 16 = 48 $ g/mol

## Step 2: Sum Atomic Masses
Molar mass ($ M $) = $ 137 + 12 + 48 = 197 \, \text{g/mol} $.
# Answer: $ 197 \, \text{g/mol} $

### Problem 2: Phosphorus Atoms in $ \text{P}_2\text{O}_5 $
# Explanation:
The formula $ \text{P}_2\text{O}_5 $ indicates 2 phosphorus (P) atoms per molecule.
# Answer: 2 phosphorus atoms

### Problem 3: Seven Diatomic Elements
# Explanation:
Diatomic elements are molecules with 2 atoms. The seven are: $ \text{H}_2, \text{O}_2, \text{N}_2, \text{Cl}_2, \text{Br}_2, \text{I}_2, \text{F}_2 $ (mnemonic: "Have No Fear Of Ice Cold Beer").
# Answer: $ \text{H}_2, \text{O}_2, \text{N}_2, \text{Cl}_2, \text{Br}_2, \text{I}_2, \text{F}_2 $

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Question

Explanation:

Problem 4: How many atoms of carbon are in 3.6 moles of carbon?

Step 1: Recall Avogadro's Number

Step 2: Substitute Values

Step 1: Molar Mass of \( \text{H}_2 \)

Step 2: Substitute Values

Step 1: STP Molar Volume

Step 2: Substitute Values

Answer:

Problem 5: Moles of hydrogen in 362.8 g of \( \text{H}_2 \)