An alloy is made of 60% aluminum and 40% tin by mass. Given the densities of aluminum and tin, calculate the density of the alloy.

The density of an alloy ($\rho_{alloy}$) composed of components with mass fractions $f_i$ and densities $\rho_i$ is given by:

$$\frac{1}{\rho_{alloy}} = \sum_{i} \frac{f_i}{\rho_i}$$

In this case, for aluminum and tin:

$$\frac{1}{\rho_{alloy}} = \frac{f_{Al}}{\rho_{Al}} + \frac{f_{Sn}}{\rho_{Sn}}$$

Where $f_{Al} = 0.60$ (60%) and $f_{Sn} = 0.40$ (40%).

$$\frac{0.60}{2700} = \frac{6}{27000} = \frac{1}{4500} \text{ m}^3\text{kg}^{-1}$$ $$\frac{0.40}{7300} = \frac{4}{73000} = \frac{1}{18250} \text{ m}^3\text{kg}^{-1}$$

$$\frac{1}{\rho_{alloy}} = \frac{1}{4500} + \frac{1}{18250}$$

To add these fractions, find a common denominator or convert to decimals.

$$\frac{1}{\rho_{alloy}} = \frac{18250}{4500 \times 18250} + \frac{4500}{4500 \times 18250}$$ $$\frac{1}{\rho_{alloy}} = \frac{18250 + 4500}{82125000} = \frac{22750}{82125000} \text{ m}^3\text{kg}^{-1}$$

$$\rho_{alloy} = \frac{82125000}{22750} \text{ kgm}^{-3}$$ $$\rho_{alloy} = 3610 \text{ kgm}^{-3}$$

The density of the alloy is $\boxed{3610 \text{ kgm}^{-3}}$.

We use the formula $M = \rho \times V$.

$$M_{alloy, total} = \rho_{alloy} \times V_{alloy, total}$$ $$M_{alloy, total} = 3610 \text{ kgm}^{-3} \times 0.02 \text{ m}^3$$ $$M_{alloy, total} = 72.2 \text{ kg}$$

The alloy is 40% tin by mass.

$$M_{Sn} = 0.40 \times M_{alloy, total}$$ $$M_{Sn} = 0.40 \times 72.2 \text{ kg}$$ $$M_{Sn} = 28.88 \text{ kg}$$

The mass of tin in $0.02 \text{ m}^3$ of the alloy is $\boxed{28.88 \text{ kg}}$.