C12H22O11   is burned in pure oxygen to give only carbon dioxide and water.?
957905
the balanced reaction\nC12H22O11 + 12 O2 ---> 12 CO2 + 11 H2O\nso the right answer are:\nc. \nand\ne.
303776
Gasoline + Oxygen = CO2 + H2O + NOx + SO4 + energy
278240
C6H12O2 + 6O2 --> 6CO2 + 6H2O\n\nis the combustion reaction for glucose.
807508
Most of the info provided is superfluous.\nSimply put, you burned something and got 4.161g H2O (MW=18.016), all H coming from the sample.\nSo you formed 4.161/18.016 = 0.23096 mols H2O, and since each mol H2O contains 2 mols H, your sample contained 0.4619 mols of H.\n\nBTW, it is easy to deduce the empirical formula of the compound:\nNo hydrocarbon can have a MW of 60.095, and since only CO2 and H2O were formed, the compound  must contain O...this works out to be C3H8O, which means it is probably propanol (n or iso) or methyl ethyl ether.
458346
2NaOH + CO2 = Na2CO3 + H2O
633274
oxygen is removed and replaced by co2
795154
2CO(g) + O2(g) --> 2CO2(g)
229945
-> 6CO2 + 6H2O     Carbon dioxide.
770748
Oxygen is the limiting reactant....you would need 5.6 moles of it for a complete reaction.\n\nMoles of CO2...2.8.  It is 1/1 with methane\n\nGrams 2.8 X 44 = 123.2 grams
336167
oxygen and carbon dioxide
890894
C12H22O11+12O2-------12CO2+11H2O+E
696429
sodium ethanoate and carbon dioxide along with water
222974
Two hydrogen atoms, one oxygen atom.  H2O
883220
H2+O2-->H2O\nNaOH+HCl-->NaCl+H2O\nKOH-->(on heating,deomposition)K2O+H2O\ni guess this is what u want\nand i f u didn't get it i am sorry......
909109
Carbon Dioxide and Water
744647
C(3)H(8)+5O(2)=>3CO(2)+4H(2)O\nfuel+oxygen=>carbon dioxide+water\ncombustion \nbrackets-subscript
929071
2C18H18 + 45 O2 --->  36 CO2 + 18 H2O\n3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O