7. If there are $$ 3.00 imes 10^{23} $$ formula units of feCliz present a. How many moles is this? b. How many grams would this be?

Question

7. If there are $$ 3.00 	imes 10^{23} $$ formula units of feCliz present a. How many moles is this? b. How many grams would this be?

UpStudy AI · Accepted Answer

Sure, let's break down the problem step by step.

**Given:**
- Number of formula units of FeCl₂: $$ 3.00 \times 10^{23} $$

We'll solve for:
a. Number of moles
b. Mass in grams

---

### **a. Calculating the Number of Moles**

**Avogadro's Number ( $$ N_A $$ ):** $$ 6.022 \times 10^{23} $$ formula units/mol

**Formula:**
$$
\text{Number of moles} = \frac{\text{Number of formula units}}{N_A}
$$

**Calculation:**
$$
\text{Number of moles} = \frac{3.00 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.498 \text{ moles}
$$

**Rounded Answer:**
$$
\approx 0.50 \text{ moles}
$$

---

### **b. Calculating the Mass in Grams**

**Step 1: Determine the Molar Mass of FeCl₂**

- **Fe (Iron):** Approximately 55.85 g/mol
- **Cl (Chlorine):** Approximately 35.45 g/mol each

Since there are two chlorine atoms in FeCl₂:
$$
\text{Molar mass of FeCl}_2 = 55.85 \text{ g/mol} + 2 \times 35.45 \text{ g/mol} = 126.75 \text{ g/mol}
$$

**Step 2: Calculate the Mass**

**Formula:**
$$
\text{Mass} = \text{Number of moles} \times \text{Molar mass}
$$

**Calculation:**
$$
\text{Mass} = 0.50 \text{ moles} \times 126.75 \text{ g/mol} = 63.375 \text{ g}
$$

**Rounded Answer:**
$$
\approx 63.4 \text{ grams}
$$

---

### **Summary of Answers**

a. **Number of moles:**  
$$\boxed{0.50\ \text{mol}}$$

b. **Mass in grams:**  
$$\boxed{63.4\ \text{g}}$$
a. 0.50 moles b. 63.4 grams

If there are formula units of feCliz present
a. How many moles is this?
b. How many grams would this be?

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If there are formula units of feCliz present a. How many moles is this? b. How many grams would this be?