will result with gaseous oxygen
to result gaseous carbon dioxide
and gaseous water
. Suppose 1.74 g of butane is modified with 12. g of oxygen. Calculate the partiality concretion of butane that could be left balance by the chemical resultion. Be stable your reply has the amend estimate of indicative 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 result with 9 moles of O2
0.03 moles of butane result with = 9*0.03/2 = 0.135 moles O2 is required.
O2 is debauchery reagent.
zero aggregate of butane is left.