Methane (1.12 liter) interacts with bromine to obtain bromomethane (1 liter). Determine the volume fraction of bromomethane

Let’s write down the solution:
1. CH4 + Br2 = CH3Br + HBr – substitution reaction, bromomethane obtained;
2. Let’s make the proportion according to the equation:
1 mol of gas at normal level – 22.4 liters;
X mol (CH4) – – 1.12 L. hence, X mol (CH4) = 1 * 1.12 / 22.4 = 0.05 mol;
Y (CH3Br) = 0.05 mol since the amount of these substances is equal to 1 mol, which confirms the equation.
3. Find the theoretical volume for the reaction product – CH3Br;
V (theoretical) = 0.05 * 22.4 = 1.12 liters.
4. Let’s calculate the volume fraction – СН3Br;
V (practical) / V (theoretical) * 100 = 1 / 1.12 * 100 = 89.28%
Answer: bromomethane was formed during the reaction, its volume fraction is 89.28%