Question

Gaseous ethane (CH,CH,) will react with gaseous oxygen (0) to produce gaseous carbon dioxide (CO) and gaseous water (HO). Suppose 15. g of ethane is mixed with 76.5 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits

Answer 1

Balanced chemical reaction is

2CH₃CH₃ + 7O₂    $//img.homeworklib.com/questions/455705b0-db94-11eb-ab9c-5d6e5886e2c8.png?x-oss-process=image/resize,w_560$ 4CO₂ + 6H₂O

Calculate limiting reactant

mass of ethane = 15 gm

mass of oxygen = 76.5 gm

molar mass of ethane = 30.07 g/mol

molar mass of O₂ = 31.9988 g/mol

no. of mole = gm of compound / molar mass

mole of ethane = 15 / 30.07 = 0.49884 mole

mole of O₂ = 76.5 / 31.9988 = 2.39 moles

According to balanced chemical eqation 2 mole ethane react with  7 mole O₂​​​​​​​ molar ratio between ethane to O₂​​​​​​​ is 2: 7 therefore to react with 2.39 mole of O₂ required ethane = 2.39 X 2 / 7 = 0.6829 mole but ethane  given only 0.49884 mole therefore ethane​​​​​​​ is limiting reactant.

According to balanced reaction 2 mole ethane​​​​​​​ produce 4 mole CO₂ molar ratio between ethane​​​​​​​  to CO₂ is 2:4 therefore 0.49884 mole of ethane​​​​​​​ produce CO₂ = 0.49884 X 4 / 2 = 0.99768 mole of CO₂

gm of compound = no. of mole X molar mass

moles of CO₂ = 0.99768 moles

molar mass of CO₂ = 44.01 g/mol

gm of aspirin = 0.99768 X 44.01 = 43.91 gm

Maximum mass of carbon dioxide that could be produced = 43.91A gm