13. $\mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{K}_{2} \mathrm{CO}_{3}=2 \mathrm{KAlO}_{2}+\mathrm{CO}_{2} \uparrow$

Formulation of the reaction equation and balancing coefficients - 4 points 14. $\mathrm{KAlO}_{2}+\mathrm{HCl}=\mathrm{AlCl}_{3}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{KCl}$

Formulation of the reaction equation and balancing coefficients - 4 points 15. $\mathrm{AlCl}_{3}+3 \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}=\mathrm{Al}(\mathrm{OH})_{3} \downarrow+3 \mathrm{NH}_{4} \mathrm{Cl}$

Formulation of the reaction equation and balancing coefficients - 4 points 16. $\mathrm{Al}(\mathrm{OH})_{3}+\mathrm{NaOH}=\mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_{4}\right]$

Formulation of the reaction equation and balancing coefficients - 4 points

## TOTAL 16 points

## TASK № 3

When 28.8 g of a primary monatomic alcohol is treated with sodium, 4.48 liters of gas (STP) is released. Determine the formula of the alcohol.

#
# Solution:

17. Let's write the general equation for the reaction of alcohols with sodium:

$2 \mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+1} \mathrm{OH}+2 \mathrm{Na}=2 \mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+1} \mathrm{ONa}+\mathrm{H}_{2} \uparrow$

Writing the reaction equation and balancing the coefficients - 2 points

18. Determine the amount of gas released

$$
\mathrm{n}\left(\mathrm{H}_{2}\right)=\frac{m}{M}=\frac{4.48}{22.4}=0.2 \text { mol }
$$

Determining the amount of substance - 2 points

19. Calculate the amount of alcohol (in moles) that reacted: $\mathrm{n}\left(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+1} \mathrm{OH}\right)=\mathrm{n}\left(\mathrm{H}_{2}\right)=0.2$ mol

Determining the amount of substance - 2 points

20. Find the molar mass of the alcohol:

$$
\mathrm{M}\left(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+1} \mathrm{OH}\right)=\mathrm{m} / \mathrm{n}=28.8 / 0.2=144
$$

Determining the molecular mass of the substance - 2 points

21. Establish the molecular formula of the alcohol:

$\mathrm{M}\left(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+1} \mathrm{OH}\right)=12 \mathrm{n}+2 \mathrm{n}+1+16+1=144$

$14 n+18=144$

$14 \mathrm{n}=126$

$\mathrm{n}=9$

The molecular formula of the alcohol is $\mathrm{C}_{9} \mathrm{H}_{19} \mathrm{OH}$

Establishing the molecular formula of the substance - 2 points

## TOTAL 10 points

## PROBLEM № 4

What is the composition of the mixture of monobromides formed during the bromination of isobutane, given that the reactivity ratio of $\mathrm{C}-\mathrm{H}$ bonds at primary, secondary, and tertiary carbon atoms is $1: 82: 1600$?

## Solution: