What volume of oxygen gas (O2) measured at 0°c and 1 atm, is needed to burn completely 1 l of propane gas (c3h8) measured under the same condition?
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use stoichometric
C3H8 + 5O2-----> 3CO2 + 4H2O
now u see for burning one mole of propane we wanna 5 mole of oxygen
one mole =22.4L
one mole/22.4=1L
6.022×10²³/22.4=2.68×10²²molecules
for burning 6.022×10²³ molecules ,3.01×10²⁴ molecules of oxygen
5moles of oxygen=112litre O2 wants
for burning 2.68 × 10²² ,13.4×10²³ molecules wanna
so 13.4×10²³/6.022×10²³=litre of O2/22.4
=3.0×10^25/6.022×10²³
=5 litre of O2 wants
C3H8 + 5O2-----> 3CO2 + 4H2O
now u see for burning one mole of propane we wanna 5 mole of oxygen
one mole =22.4L
one mole/22.4=1L
6.022×10²³/22.4=2.68×10²²molecules
for burning 6.022×10²³ molecules ,3.01×10²⁴ molecules of oxygen
5moles of oxygen=112litre O2 wants
for burning 2.68 × 10²² ,13.4×10²³ molecules wanna
so 13.4×10²³/6.022×10²³=litre of O2/22.4
=3.0×10^25/6.022×10²³
=5 litre of O2 wants
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