Question

Consider this reaction: 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

(a) How many grams of oxygen are required to completely react with (burn) 100.0 g of ethane?

(b) If 55.0 g of ethane is mixed with 55.0 gram of oxygen, and the mixture then reacts, how many grams of water will be formed?

Answer 1

The balanced chemical reaction is -

2C2H6 + 7 O2 → 4 CO2 + 6 H2O

molar mass of ethane = molar mass of C*2 + molar mass of H*6 = 12g/mole*2 + 1g/mole*6 = 30g/mole

no. of moles of ethane in 100g ethane = weight/molar mass = 100g / 30g/mole = 3.33 moles

From balanced chemical reaction we know that, 2 moles of ethane react with 7 moles of O2 to produces products

So, 1 mole of ethane reacts with 7/2 moles of O2 so, 3.33 moles of ethane will react with 3.33*7/2 moles of O2 = 11.67 moles

So, weight of 11.67 moles of O2 = moles*molar mass = 11.67moles*32g/mole = 373.33 g

b)

moles of ethane in 55g = weight/molar mass = 55g/ 30g/mole = 1.83 moles

moles of oxygen in 55g = weight/molar mass = 55g/ 32g/mole = 1.719 moles

But 1.83 moles of ethane will need 1.83*7/2 = 6.4 moles of O2 so, here we have O2 in limited amount

1 mole of O2 reacts with 2/7 moles of ethane so, 1.719 moles of O2 will react with 1.719*2/7 moles of ethane = 0.491 moles of ethane

From balanced chemical reaction we have the ratio of moles of ethane and water = 2/6 = 1/3 so, it means each mole of ethane when reacts give 3 moles of H2O so,

0.491 moles of ethane will give 0.491*3 moles of H2O = 1.47 moles of H2O

Molar mass of H2O = 18g/mole

weight of 1.47 moles of H2O = moles*molar mass = 1.47g/mole*18 = 26.46 g