Question

$A +2 B +3 C \rightleftharpoons AB _{2} C _{3}$ Reaction of $6.0\, g$ of $A , 6.0 \times 10^{23}$ 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 no. $=6 \times 10^{23}$ ):

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

Answer 1

The Correct Option is C

To solve this question, we need to determine the atomic mass of element \(B\) in the reaction \(A + 2B + 3C \rightleftharpoons AB _{2} C _{3}\). The following information is provided:

• Mass of \(A\) is \(6.0\, \text{g}\), and its atomic mass is \(60\, \text{amu}\).
• Atoms of \(B\) are \(6.0 \times 10^{23}\).
• Moles of \(C\) are \(0.036\, \text{mol}\), and its atomic mass is \(80\, \text{amu}\).
• Mass of the product \(AB_2C_3\) is \(4.8\, \text{g}\).

First, calculate the moles of each reactant:

1. Moles of \(A\):
   Using the formula for converting mass to moles: \(\text{Moles of } A = \frac{\text{Mass of } A}{\text{Atomic mass of } A} = \frac{6\, \text{g}}{60\, \text{g/mol}} = 0.1\, \text{mol}\)
2. Moles of \(B\):
   As per Avogadro's number, \(6.0 \times 10^{23}\) atoms of \(B\) is equivalent to: \(\text{Moles of } B = \frac{6.0 \times 10^{23}}{6.0 \times 10^{23}} = 1\, \text{mol}\)
3. Moles of \(C\):
   Given directly as \(0.036\, \text{mol}\).

Since the reaction is:

\(A + 2B + 3C \rightleftharpoons AB_2C_3\)

Product Yield Analysis:

The reaction yields \(4.8\, \text{g}\) of \(AB_2C_3\).

The molar mass of \(AB_2C_3\) can be expressed using the atomic masses of \(A\), \(B\), and \(C\):

\(\text{Molar mass of } AB_2C_3 = M_A + 2M_B + 3M_C\)

Substituting known values, we get:

\(= 60 + 2M_B + 3\times 80\)

\(= 60 + 2M_B + 240\)

\(= 300 + 2M_B\)

Given that \(AB_2C_3\) has a mass of \(4.8\, \text{g}\), and assuming \(0.036\, \text{mol}\) (from \(C\)) was used completely, then:

\(\text{Mass of } AB_2C_3 \text{ per mole} = \frac{4.8\, \text{g}}{0.036\, \text{mol}}\)

300 + 2M_B = 133.33

Solving for \(M_B\):

\(2M_B = 133.33 - 300\)
\(2M_B = -166.67\)
\(M_B = \frac{-166.67}{2} = 50\)

Therefore, the atomic mass of \(B\) is 50 amu.