Question

3. Given the following reaction: 2C₂H6(l) + 7 O₂(g) ---- 4CO₂(g) + 6H₂O(l)

How many moles of O2 is needed to produce 5.00 g H2O?

If 8.32 g of ethane react, how many grams of carbon dioxide will be formed? How many grams of water will be produced?

Answer 1

Number of moles of H2O = 5.00 g / 18.0 g.mol^-1 = 0.278 mole

from the balanced equation we can say that

6 mole of H2O is produced by 7 mole of O2 so

0.278 mole of H2O will be produced by

= 0.278 mole of H2O *(7 mole of O2 / 6 mole of H2O )

= 0.324 mole of O2

Therefore, the number of moles of O2 produced will be 0.324

Number of moles of ethane = 8.32 g/ 30.07 g.mol^-1 = 0.277 mole

from the balanced equation we can say that

2 mole of ethane produces 4 mole of CO2 so

0.277 mole of ethane will produce 0.554 mole of CO2

1 mole of CO2 = 44.01 g

so 0.554mole of CO2 = 24.4 g

Therefore, the mass of CO2 produced will be 24.4 g

from the balanced equation we can say that

2 mole of ethane produces 6 mole of H2O so

0.277 mole of ethane will produce 0.831 mole of H2O

1 mole of H2O = 18 g

so 0.831 mole of H2O = 14.95 g

Therefore, the mass of H2O produced will be 14.95 g