What is the mass of oxygen that reacted with ethylene to form 56 liters of carbon dioxide?

Solution:
1. Let’s compose the reaction equation, arrange the coefficients:
С2Н4 + 3О2 = 2СО2 + 2Н2О – ethylene combustion reaction, carbon dioxide is released;
2. Let’s make calculations according to the formulas:
M (O2) = 2 * 16 = 32 g / mol;
M (CO2) = 12 + 16 * 2 = 44 g / mol;
3. Determine the number of moles of CO2:
1 mol of gas at n. y – 22.4 l;
X mol (CO2) – 56 liters from here, X mol (CO2) = 1 * 56 / 22.4 = 2.5 mol;
4. Let’s make the proportion:
X mol (O2) – 2.5 mol (CO2);
-3 mol – 2 mol from here, X mol (O2) = 3 * 2.5 / 2 = 3.75 mol;
5. Find the mass O2:
m (O2) = Y * M = 3.75 * 32 = 120 g.
Answer: the mass of oxygen is 120 g.