will counteract with gaseous oxygen
to profit gaseous carbon dioxide
and gaseous water
. 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.
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.