In the following reaction:
$$A + 2B + 3C \rightleftharpoons AB_2C_3$$
6.0 g of A, $$6.0 \times 10^{23}$$ atoms of B and 0.036 mol of C reacted and formed 4.8 g of compound $$AB_2C_3$$. If the atomic mass of A and C are 60 and 80 amu, respectively. What is the atomic mass of B in amu? (Avogadro number = $$6 \times 10^{23}$$)

We have the balanced chemical equation

$$A + 2B + 3C \;\rightleftharpoons\; AB_2C_3$$

First we convert every given quantity of the reactants into moles, because the stoichiometric coefficients in the equation relate moles to moles.

For element A the atomic (molar) mass is given as 60 amu, so by the definition of mole

$$n_A \;=\;\frac{\text{mass of }A}{\text{molar mass of }A} \;=\;\frac{6.0\;\text{g}}{60\;\text{g mol}^{-1}} =0.10\;\text{mol}.$$

For element B we are told the number of atoms directly: $$6.0\times10^{23}$$ atoms. One mole contains Avogadro’s number, $$N_A = 6\times10^{23}$$ atoms, so

$$n_B \;=\;\frac{6.0\times10^{23}\;\text{atoms}}{6\times10^{23}\;\text{atoms mol}^{-1}} =1.0\;\text{mol}.$$

For element C the amount is already supplied as moles:

$$n_C = 0.036\;\text{mol}.$$

Now we check which reactant limits the formation of product. According to the equation, to make 1 mol of $$AB_2C_3$$ we must consume 1 mol of A, 2 mol of B and 3 mol of C. Using the available moles of each reactant we calculate the maximum moles of product each could furnish:

From A: $$\dfrac{0.10\;\text{mol A}}{1\;\text{mol A per mol product}} = 0.10\;\text{mol product}.$$

From B: $$\dfrac{1.0\;\text{mol B}}{2\;\text{mol B per mol product}} = 0.50\;\text{mol product}.$$

From C: $$\dfrac{0.036\;\text{mol C}}{3\;\text{mol C per mol product}} = 0.012\;\text{mol product}.$$

The smallest of these is $$0.012\;\text{mol}$$, so C is the limiting reactant and the reaction can form at most $$0.012\;\text{mol}$$ of $$AB_2C_3$$.

The experiment actually produces 4.8 g of the compound. Therefore the number of moles of product that were formed is

$$n_{\text{prod}} \;=\;\frac{4.8\;\text{g}}{M_{\text{prod}}},$$

where $$M_{\text{prod}}$$ is the molar mass of $$AB_2C_3$$. Because only 0.012 mol can exist (C is limiting), we equate

$$\frac{4.8}{M_{\text{prod}}} = 0.012.$$

Next we express $$M_{\text{prod}}$$ in terms of the unknown atomic mass of B, which we call $$x\;\text{amu}$$. The product contains one atom of A, two atoms of B and three atoms of C, so

$$M_{\text{prod}} = M_A + 2M_B + 3M_C = 60 + 2x + 3(80).$$

Calculating the constant part,

$$3(80) = 240,$$

hence

$$M_{\text{prod}} = 60 + 2x + 240 = 300 + 2x.$$

We now substitute this into the earlier mole equation:

$$\frac{4.8}{300 + 2x} = 0.012.$$

Cross-multiplying gives

$$4.8 = 0.012\,(300 + 2x).$$

Dividing the right-hand side coefficient first,

$$\frac{4.8}{0.012} = 300 + 2x.$$

Since $$\dfrac{4.8}{0.012} = 400,$$ we have

$$400 = 300 + 2x.$$

Subtracting 300 from both sides yields

$$2x = 100,$$

so

$$x = 50.$$

This value of $$x$$ is the atomic mass of element B, expressed in atomic mass units (amu).

Hence, the correct answer is Option D.