In the above reaction,5 g of toluene is converted into benzaldehyde with 92% yield. The amount of benzaldehyde produced is _____ $$\times 10^{-2}$$ g

Toluene:

$$\mathrm{C_7H_8}$$

Molar mass of toluene:

$$\mathrm{= (7\times12) + (8\times1)}$$

$$\mathrm{= 84 + 8 = 92\ g\ mol^{-1}}$$

Benzaldehyde:

$$\mathrm{C_7H_6O}$$

Molar mass of benzaldehyde:

$$\mathrm{= (7\times12) + (6\times1) + 16}$$

$$\mathrm{= 84 + 6 + 16 = 106\ g\ mol^{-1}}$$

Moles of toluene used:

$$\mathrm{= \frac{5}{92}\ mol}$$

Reaction stoichiometry:

$$\mathrm{1\ mol\ Toluene \rightarrow 1\ mol\ Benzaldehyde}$$

Therefore, theoretical moles of benzaldehyde:

$$\mathrm{= \frac{5}{92}\ mol}$$

Theoretical mass of benzaldehyde:

$$\mathrm{= \frac{5}{92}\times106\ g}$$

Given yield:

$$\mathrm{92\%}$$

Actual mass produced:

$$\mathrm{= \left(\frac{5}{92}\times106\right)\times\frac{92}{100}}$$

$$\mathrm{= \frac{5\times106}{100}}$$

$$\mathrm{= \frac{530}{100}}$$

$$\mathrm{= 5.3\ g}$$

Expressing in the form $$\mathrm{\times10^{-2}\ g}$$:

$$\mathrm{5.3\ g = 530\times10^{-2}\ g}$$

$$\boxed{\mathrm{530}}$$