3. how many grams are in of 4.50 moles of barium sulfide?
4. how many formula units are there in 5.39 moles of silver chloride?
two - step problems:
5. how many formula units are in 937 g of calcium phosphate?
6. how many grams are in 3.21×10²⁴ formula units of potassium hydroxide?
7. how many formula units are in 2.00 grams of aluminum fluoride?
8. how many molecules are in 654.5 grams of sulfur hexafluoride?
tricky problems:
9. how many atoms of oxygen are in 2.5 moles of oxygen gas?
10. how many grams of chlorine are in 5.55×10²⁶ formula units of magnesium chloride?

Question

Accepted Answer

### Problem 3: How many grams are in 4.50 moles of barium sulfide?
# Explanation:
## Step 1: Determine the molar mass of BaS
Barium (Ba) has a molar mass of $ 137.33 \, \text{g/mol} $, sulfur (S) has a molar mass of $ 32.07 \, \text{g/mol} $. So, molar mass of $ \text{BaS} = 137.33 + 32.07 = 169.40 \, \text{g/mol} $.
## Step 2: Calculate mass from moles
Use the formula $ \text{mass} = \text{moles} \times \text{molar mass} $. Substitute moles $ = 4.50 \, \text{mol} $ and molar mass $ = 169.40 \, \text{g/mol} $.
$ \text{mass} = 4.50 \, \text{mol} \times 169.40 \, \text{g/mol} $
# Answer:
$ 762.3 \, \text{grams} $ (or $ 762 \, \text{g} $ if rounded to three significant figures)

### Problem 4: How many formula units are there in 5.39 moles of silver chloride?
# Explanation:
## Step 1: Recall Avogadro's number
Avogadro's number is $ 6.022 \times 10^{23} \, \text{formula units/mol} $.
## Step 2: Calculate formula units
Use the formula $ \text{formula units} = \text{moles} \times \text{Avogadro's number} $. Substitute moles $ = 5.39 \, \text{mol} $.
$ \text{formula units} = 5.39 \, \text{mol} \times 6.022 \times 10^{23} \, \text{formula units/mol} $
# Answer:
$ 3.24 \times 10^{24} \, \text{formula units} $

### Problem 5: How many formula units are in 937 g of calcium phosphate?
# Explanation:
## Step 1: Molar mass of $ \text{Ca}_3(\text{PO}_4)_2 $
Calcium (Ca): $ 40.08 \, \text{g/mol} $, phosphorus (P): $ 30.97 \, \text{g/mol} $, oxygen (O): $ 16.00 \, \text{g/mol} $.
Molar mass $ = 3(40.08) + 2(30.97) + 8(16.00) = 120.24 + 61.94 + 128.00 = 310.18 \, \text{g/mol} $.
## Step 2: Moles of $ \text{Ca}_3(\text{PO}_4)_2 $
$ \text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{937 \, \text{g}}{310.18 \, \text{g/mol}} \approx 3.021 \, \text{mol} $.
## Step 3: Formula units
Use Avogadro's number: $ \text{formula units} = \text{moles} \times 6.022 \times 10^{23} $.
$ \text{formula units} = 3.021 \, \text{mol} \times 6.022 \times 10^{23} \, \text{formula units/mol} \approx 1.82 \times 10^{24} $.
# Answer:
$ \approx 1.82 \times 10^{24} \, \text{formula units} $

### Problem 6: How many grams are in $ 3.21 \times 10^{24} $ formula units of potassium hydroxide?
# Explanation:
## Step 1: Moles from formula units
Use $ \text{moles} = \frac{\text{formula units}}{\text{Avogadro's number}} $. Substitute formula units $ = 3.21 \times 10^{24} $, Avogadro's number $ = 6.022 \times 10^{23} $.
$ \text{moles} = \frac{3.21 \times 10^{24}}{6.022 \times 10^{23}} \approx 5.33 \, \text{mol} $.
## Step 2: Molar mass of KOH
Potassium (K): $ 39.10 \, \text{g/mol} $, oxygen (O): $ 16.00 \, \text{g/mol} $, hydrogen (H): $ 1.01 \, \text{g/mol} $.
Molar mass $ = 39.10 + 16.00 + 1.01 = 56.11 \, \text{g/mol} $.
## Step 3: Mass from moles
$ \text{mass} = \text{moles} \times \text{molar mass} = 5.33 \, \text{mol} \times 56.11 \, \text{g/mol} \approx 299 \, \text{g} $.
# Answer:
$ \approx 299 \, \text{grams} $

### Problem 7: How many formula units are in 2.00 grams of aluminum fluoride?
# Explanation:
## Step 1: Molar mass of $ \text{AlF}_3 $
Aluminum (Al): $ 26.98 \, \text{g/mol} $, fluorine (F): $ 19.00 \, \text{g/mol} $.
Molar mass $ = 26.98 + 3(19.00) = 26.98 + 57.00 = 83.98 \, \text{g/mol} $.
## Step 2: Moles of $ \text{AlF}_3 $
$ \text{moles} = \frac{2.00 \, \text{g}}{83.98 \, \text{g/mol}} \approx 0.0238 \, \text{mol} $.
## Step 3: Formula units
$ \text{formula units} = \text{moles} \times 6.022 \times 10^{23} = 0.0238 \, \text{mol} \times 6.022 \times 10^{23} \, \text{formula units/mol} \approx 1.43 \times 10^{22} $.
# Answer:
$ \approx 1.43 \times 10^{22} \, \text{formula units} $

### Problem 8: How many molecules are in 654.5 grams of sulfur hexafluoride?
# Explanation:
## Step 1: Molar mass of $ \text{SF}_6 $
Sulfur (S): $ 32.07 \, \text{g/mol} $, fluorine (F): $ 19.00 \, \text{g/mol} $.
Molar mass $ = 32.07 + 6(19.00) = 32.07 + 114.00 = 146.07 \, \text{g/mol} $.
## Step 2: Moles of $ \text{SF}_6 $
$ \text{moles} = \frac{654.5 \, \text{g}}{146.07 \, \text{g/mol}} \approx 4.481 \, \text{mol} $.
## Step 3: Molecules
$ \text{molecules} = \text{moles} \times 6.022 \times 10^{23} = 4.481 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 2.699 \times 10^{24} $.
# Answer:
$ \approx 2.70 \times 10^{24} \, \text{molecules} $

### Problem 9: How many atoms of oxygen are in 2.5 moles of oxygen gas?
# Explanation:
## Step 1: Recognize $ \text{O}_2 $ is diatomic
Oxygen gas is $ \text{O}_2 $, so 1 mole of $ \text{O}_2 $ has 2 moles of O atoms.
## Step 2: Moles of O atoms
Moles of O atoms $ = 2.5 \, \text{mol} \, \text{O}_2 \times 2 = 5.0 \, \text{mol} \, \text{O} $.
## Step 3: Atoms of O
$ \text{atoms} = \text{moles} \times 6.022 \times 10^{23} = 5.0 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 3.011 \times 10^{24} $.
# Answer:
$ \approx 3.0 \times 10^{24} \, \text{atoms} $ (or $ 3.01 \times 10^{24} $)

### Problem 10: How many grams of chlorine are in $ 5.55 \times 10^{26} $ formula units of magnesium chloride?
# Explanation:
## Step 1: Moles of $ \text{MgCl}_2 $
$ \text{moles} = \frac{\text{formula units}}{\text{Avogadro's number}} = \frac{5.55 \times 10^{26}}{6.022 \times 10^{23}} \approx 921.6 \, \text{mol} $.
## Step 2: Moles of Cl in $ \text{MgCl}_2 $
1 mole of $ \text{MgCl}_2 $ has 2 moles of Cl. So, moles of Cl $ = 921.6 \, \text{mol} \times 2 = 1843.2 \, \text{mol} $.
## Step 3: Mass of Cl
Molar mass of Cl is $ 35.45 \, \text{g/mol} $.
$ \text{mass} = 1843.2 \, \text{mol} \times 35.45 \, \text{g/mol} \approx 6.53 \times 10^{4} \, \text{g} $ (or 65300 g).
# Answer:
$ \approx 6.53 \times 10^{4} \, \text{grams} $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 3: How many grams are in 4.50 moles of barium sulfide?

Step 1: Determine the molar mass of BaS

Step 2: Calculate mass from moles

Step 1: Recall Avogadro's number

Step 2: Calculate formula units

Step 1: Recall Avogadro's number

Step 2: Calculate formula units

Answer:

Problem 4: How many formula units are there in 5.39 moles of silver chloride?