22 grams carbon dioxide (0.5 moles)
64 grams
If all of the quantities stated actually reacted, the law of the conservation of mass shows that the mass of carbon dioxide produced would be 40 - 18 or 22 grams.
The mass of carbon dioxide is 141,2 g.
114,426 g carbon dioxide are produced.
If 15 liters of propane are completely consumed 90,25 grams of carbon dioxide are produced.
The gram molecular mass of hexane is 86.18. Therefore, 25.0 g of hexane constitute 25.0/86.18 or 0.290 moles. Each mole of hexane contains six carbon atoms and therefore will produce six molecules of carbon dioxide by burning in an excess of oxygen. 6 X 0.290 = 1.74 moles of carbon dioxide. The gram molecular mass of carbon dioxide is 44.00. Therefore, the mass of carbon dioxide produced will be 1.74 X 44.00 or 76.6 grams of carbon dioxide, to the justified number of significant digits.
11 grams because all is reacted and there is no reactant left over, although if there were only 3 grams of carbon there would have to be 6 grams of oxygen for this to be viable as carbon dioxide is CO2 so the question asked was itself wrong.
66g
If all of the quantities stated actually reacted, the law of the conservation of mass shows that the mass of carbon dioxide produced would be 40 - 18 or 22 grams.
888
The equation for the reaction is C + O2 -> CO2. The relevant gram atomic masses are 12.011 for carbon and 15.9994 for oxygen. Therefore, the ratio of the mass of carbon dioxide produced to carbon burnt is [2(15.9994) + 12.011]/12.011 or about 3.66. From burning 3 grams of carbon, the mass of carbon dioxide produced is therefore 1 X 101 grams, to the justified number of significant digits.
The mass of carbon dioxide is 141,2 g.
114,426 g carbon dioxide are produced.
51.4 - 51.8
If 15 liters of propane are completely consumed 90,25 grams of carbon dioxide are produced.
C6H12O6 + 6O2 --> 6CO2 + 6H2O 45 grams C6H12O6 (1 mole C6H12O6/180.156 grams)(6 moles CO2/1 mole C6H12O6)(44.01 grams/1 mole CO2) = 66 grams carbon dioxide produced ==========================
62.1 - 62.5
Balanced equation always and first. Decomposition reaction. CO2 -> C + O2 440 grams CO2 (1 mole CO2/44.01 grams)(1 mole O2/1 mole CO2)(32 grams/1 mole O2) = 319.93 grams O2 ( call it 320 grams )