Answer: 34.4 g
Explanation:


As
is in excess,
is the limiting reagent and thus it will limit the formation of products.

According to stoichiometry:
2 moles of hydrogen produce = 1 mole of 
2.5 moles of hydrogen produce =
of 
Mass of 
But as % yield is 86%, mass of
produded is 
Thus 34.4 g of
is produced.
Answer:
435.38 L
Explanation:
From the question given above, the following data were obtained:
Initial mole (n₁) = 3.25 mole
Initial volume (V₁) = 100 L
Final mole (n₂) = 14.15 mole
Final volume (V₂) =?
The final volume occupied by the gas can be obtained as follow:
V₁/n₁ = V₂/n₂
100 / 3.25 = V₂ / 14.15
Cross multiply
3.25 × V₂ = 100 × 14.15
3.25 × V₂ = 1415
Divide both side by 3.25
V₂ = 1415 / 3.25
V₂ = 435.38 L
Thus, the final volume of the gas is 435.38 L
Answer:
The type of reaction for the following equation is combustion equation.
Explanation:
Combustion reaction is defined as the chemical reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.

The reaction given to us:

When 1 mole of ethane reacts with 7/2 moles of oxygen gas it gives 3 moles of water and 2 moles of carbon dioxide gas.
The type of reaction for the following equation is combustion equation.
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