How much ethane can be obtained by reacting 284 grams of chloromethane with 50 grams of sodium?

| June 11, 2021 | education

Let’s implement the solution:
1. We write down the equation according to the problem statement:
m = 284 g. m = 50 g. X l. -?
2CH3Cl + 2Na = C2H6 + 2NaCl – ethane obtained by the Wurtz reaction;
2. We make calculations:
M (CH3Cl) = 50.4 g / mol;
M (Na) = 22.9 g / mol.
3. Determine the number of moles of chloromethane and sodium, if the mass is known:
Y (Na) = m / M = 50 / 22.9 = 2.18 mol (deficient substance);
Y (CH3Cl) = m / M = 284 / 50.4 = 5.6 mol (substance in excess);
Calculations are made for the substance in deficiency.
4. Proportion:
2.18 mol (Na) – X mol (C2H6);
-2 mol – 1 mol from here, X mol (C2H6) = 2.18 * 1/2 = 1.09 mol.
5. Find the volume of the product:
V (C2H6) = 1.09 * 22.4 = 24.42 l.
Answer: in the course of the reaction, ethane with a volume of 24.42 liters was obtained.



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