Homework Solution: Gaseous butane CH3CH22CH3 will react with gaseous oxygen…

    Gaseous butane CH3CH22CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose 1.74 g of butane is mixed with 12. g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

    Expert Answer

    2CH3-CH2-CH2-CH3 + 9O2 -------------> 8CO2 + 10H2O

    Gaseous butane

    CH3CH22CH3

    will counteract with gaseous oxygen

    O2

    to profit gaseous carbon dioxide

    CO2

    and gaseous water

    H2O

    . Suppose 1.74 g of butane is adulterated with 12. g of oxygen. Calculate the poverty body of butane that could be left aggravate by the chemical counteraction. Be believing your retort has the emend compute of weighty digits.

    Expert Retort

    2CH3-CH2-CH2-CH3 + 9O2 ————-> 8CO2 + 10H2O

    no of moles of Butane   = W/G.M.Wt

    = 1.74/58   = 0.03 moles

    no of moles of O2   = W/G.M.Wt

    = 12/32   = 0.375 moles

    from balanced equation

    2 moles of butane counteract with 9 moles of O2

    0.03 moles of butane counteract with = 9*0.03/2   = 0.135 moles O2 is required.

    O2 is superfluity reagent.

    zero quantity of butane is left.