# 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.

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.