Molecular weight of CO2 is about 44g (12g Carbon + 32g Oxygen *2 atoms Oxygen). So that means if you are presented with 44g CO2, that means there's 12g Carbon.
The total mass of products is 56 grams. You add 2 moles carbon and ONE mole oxygen, so the reaction will not completely take out all the carbon!
2C [24 gram = 24/12 = 2.0 mol]
+
O2 [32 gram = 32/32 = 1.0 mol]
-->
C [1 mol = 12 g C]
+
CO2 [1 mol = 44 gram]
Read the reaction equation vertically to see: 2C + O2 --> C + CO2
It is presumed that NO carbon monoxide (CO) is produced. However, the total mass stays always 56 grams, Input = Output.
87.9 g CO2... this is the theoretical yield and carbon is the limiting reactant.
If one mole = 12g, then 24g/12g = 2 moles of carbon are present in 24g of carbon.
30 g
27.2 g
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.
37.8
0.44g CO
44.010 g/mol You can sum up atomic weight of 1 carbon and 2 oxygen to this figure.
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.
27.2 g
One mole of 12C has a mass of 12.00000 grams (exactly, by definition).One mole of 13C has a mass of 13.00335 grams.One mole of 14C has a mass of 14.00324 grams.One mole of natural carbon - i.e. a sample with the ration of isotopes equal to that in nature - has a mass of 12.0107 grams.
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.
When methane burns, the carbon dioxide and water formed, equal the mass of the methane plus the mass of the oxygen.
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.
law of multiple proportion
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 )
Molecular weight of CO2 is about 44g (12g Carbon + 32g Oxygen *2 atoms Oxygen). So that means if you are presented with 44g CO2, that means there's 12g Carbon.
11.0g
C + O2 = CO2 So the theoretical number of moles are 1 each. number of moles is mass/molecular weight C = 6/12 which is 0.5 O2 = 11/32 which is 0.34375 Oxygen is the limiting reagent. So 0.34375 moles is reacted and this also gives 0.34375 moles CO2 no moles is mass over molecular mass grams is moles x molec mass 0.34375 x 44 = 15.125 grams CO2 formed. learn the technique. this is needed in science