`A + 2B+ 3C AB_2 C_3`
Reaction of 6.0 g of A, `6.0 xx 10^(23)` atoms of B, and 0.036 mol of C yields 4.8 g of compound `AB_(2)C_(3)`. If the atomic mass of A and C are 60 and 80 amu, respectively. The atomic mass of B is: (Avogadro number `= 6 xx 10^(23)`)

Options

A. 70 amu
B. 60 amu
C. 50 amu
D. 40 amu

Explanation

To solve the problem step by step, we will follow the stoichiometric principles and calculations based on the information provided.

### Step 1: Determine the number of moles of A
The number of moles of a substance can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass (g)}}{\text{Molar mass (g/mol)}} \]
For substance A:
- Given mass of A = 6.0 g
- Molar mass of A = 60 g/mol

Calculating the number of moles of A:
\[ \text{Number of moles of A} = \frac{6.0 \, \text{g}}{60 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 2: Determine the number of moles of B
We are given the number of atoms of B:
\[ \text{Number of atoms of B} = 6.0 \times 10^{23} \, \text{atoms} \]
Using Avogadro's number (6.0 x 10^23 atoms/mol), we can find the number of moles of B:
\[ \text{Number of moles of B} = \frac{6.0 \times 10^{23} \, \text{atoms}}{6.0 \times 10^{23} \, \text{atoms/mol}} = 1 \, \text{mol} \]

### Step 3: Determine the number of moles of C
The number of moles of C is given directly:
\[ \text{Number of moles of C} = 0.036 \, \text{mol} \]

### Step 4: Identify the limiting reagent
The balanced chemical equation is:
\[ A + 2B + 3C \rightarrow AB_2C_3 \]
From the stoichiometry of the reaction:
- 1 mole of A reacts with 2 moles of B and 3 moles of C.

We have:
- 0.1 mol of A
- 1 mol of B (which can provide enough for 0.5 mol of A)
- 0.036 mol of C (which can provide enough for 0.012 mol of A)

Since C is the limiting reagent (as it can only react with 0.012 mol of A), we will use the moles of C to determine the amount of product formed.

### Step 5: Calculate the moles of AB2C3 formed
From the limiting reagent (C):
\[ \text{Moles of } AB_2C_3 = \frac{\text{Moles of C}}{3} = \frac{0.036}{3} = 0.012 \, \text{mol} \]

### Step 6: Calculate the molar mass of AB2C3
The mass of the compound formed is given as 4.8 g. The molar mass can be calculated using:
\[ \text{Molar mass} = \frac{\text{Mass}}{\text{Moles}} = \frac{4.8 \, \text{g}}{0.012 \, \text{mol}} = 400 \, \text{g/mol} \]

### Step 7: Set up the equation for molar mass
The molar mass of the compound can be expressed as:
\[ \text{Molar mass of } AB_2C_3 = \text{Molar mass of A} + 2 \times \text{Molar mass of B} + 3 \times \text{Molar mass of C} \]
Substituting the known values:
\[ 400 = 60 + 2x + 3 \times 80 \]
Where \( x \) is the molar mass of B.

### Step 8: Solve for x (the molar mass of B)
Calculating the right side:
\[ 400 = 60 + 2x + 240 \]
\[ 400 = 300 + 2x \]
\[ 100 = 2x \]
\[ x = 50 \, \text{g/mol} \]

### Conclusion
The atomic mass of B is 50 amu.

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