Chlorination of 2016 ml of methane consumed 0.10 mol of chlorine. Determine the amount (mol) of chlorination products.

| June 25, 2021 | education

Let’s implement the solution:
1. According to the condition of the problem, we compose the equation:
V = 2016 ml. Y = 0.10 mol. X mole -? X mole -?
CH4 + Cl2 = CH3Cl + HCl – substitutions, methane chloride, hydrogen chloride were obtained;
1 mol 1 mol 1 mol.
2. We translate the units of volume:
1 ml = 0.001 l;
Y (CH4) = 2.016 l.
3. Proportion:
1 mol of gas at normal level – 22.4 l;
X mol (CH4) – 2.016 L. hence, X mol (CH4) = 1 * 2.016 / 22.4 = 0.09 mol;
Y (CH3Cl) = 0.09 mol;
Y (HCl) = 0.09 mol since the amount of these substances according to the equation is 1 mol.
4. Find the value of the volume of products:
Y (HCl) = 0.09 * 22.4 = 2.016 L;
Y (CH3Cl) = 0.09 * 22.4 = 2.016 L.
Answer: methane chloride, hydrogen chloride with a volume of 2.016 liters are released.



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