Yo sup??
1 lmole of O2 contains 22.4 litre
therefore 10.2 miles will contain
=10.2*22.4
=228.48 litre
Hence the correct answer is option D
Hope this helps
Answer:
Since ethane is the limiting reactant. It will completely be consumed. There will remain 0 grams of ethane.
Explanation:
Step 1: Data given
Mass of C2H6 = 20.1 grams
Mass of O2 = 95.0 grams
Molar mass ethane = 30.07 g/mol
Molar mass O2 = 32.0 g/mol
Molar mass CO2 = 44.01 g/mol
Step 2: The balanced equation
2C2H6 + 7O2 → 4CO2 + 6H2O
Step 3: Calculate moles C2H6
Moles C2H6 = mass C2H6 / molar mass C2H6
Moles C2H6 = 20.1 grams /30.07 g/mol
Moles C2H6 = 0.668 moles
Step 4: Calculate moles O2
Moles O2 = 95.0 grams / 32.0 g/mol
Moles O2 = 2.97 moles
Step 5: Calculate limiting reactant
For 2 moles C2H6 we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O
C2H6 is the limiting reactant. It will completely be consumed ( 0.668 moles).
O2 is in excess. There will react 7/2 * 0.668 = 2.338 moles
There will remain 2.97 - 2.338 = 0.632 moles O2
Since ethane is the limiting reactant. It will completely be consumed. There will remain 0 grams of ethane.
Answer:
D
Explanation:
B is solid and liquid while D is liquid and gas
Answer: hydrogen has 0.0504% and Fluorine has 0.950%
Explanation: H= 1.00794
F= 18.9984
1 X 1.00794+ 1 X 18.9984= 20.00778g/mol
H= 1.00794/20.00778= 0.0504%
F-18.9984/20.00778= 0.950%