Question

12. If 4.95 g of ethylene (C2H4) are combusted with 3.25 g of oxygen. a. What is the limiting reagent? b. How many grams of CO2 are formed? 1 LIITI . . . .. . .

Answer 1

At first we have to know about the limitting reagent. what is a limiting reagent?? - limitting reagent 5 the that type of substance which is conensed totally when the reaction is completed. The product of the chemical reaction depends on the limitting reagent. Here CaHy reacts with O2. The balanced Chemical equation for this reaction is a Caty + 30₂ (9) 26026) + 2H20 (9) so from this reaction we know that 1 mol of CH4 reacts with 3 mol of O2 to produce 2 mol of coz and 2 mol of Ho. The molecular weight of ₂ Hy = 28 g/mol in = (16 x2) - 329m/mol so, according to the question - 4.95 gm of CH4 = 4.95 = 0.177 mol of Ca Hy

O2 3.25 gm 3. 25 3. 25 of O2 = 31082) = 32 = 0.102 mol of Now The reaction to be completed - for the complete combution of 1 mol of Catty , 3 mol of Oz is required. for combusion of o. 177 mol of CH4 , amount of required O2 - (340.177) = 0:531 mot. But the O2 present i 0.102 mol which is in less amount than the

required amount. so, Oq ý the limmitting The amount of product depend on the amount reagent. reagent. formed is of limitting i for from - 3 mol of Oz , 2 mol of Coz is produced coz is produced. 0.102 » 02 (230;102 ) = 0.068 mol. : 0068 mol of con would produced.